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Thread: Group Theory for We Beginners

  1. #1 Group Theory for We Beginners 
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    Because I get bored with calculus and New Age thinking, I shall continue my quest to bring my (limited version of) abstract mathematics to this forum. And the first person to say "get a life" will have lost theirs (only teasing!). But groups are at one and the same intuititive and surprising. Let's see.

    First let's have a definition:
    A group G is a set which has
    a well-defined, associative and closed operation
    an identity e
    an inverse.

    Let's unpack this. In group theory, the group operation is always referred to as "group multiplication". But don't be fooled: group theorists don't always mean multplication in the arithmetic sense; it might mean addition. There is a convention which I'll explain later.

    The operation, group multiplication, is said to be "closed" if the result of the operation is still an element of the group (are the primes then a group?).

    The identity, usually written e, is 0 when the operation is (arithmetic) addition, 1 when it is (arithmetic) multiplication. The inverses are respectively -x and 1/x.

    Now some notational conventions. Some authors use the centre dot to denote the group operation thus a·b, but most use juxtaposition thus ab. Don't assume this means arithmetic product, though, we might be in
    an additive group.

    Similarly most books use x<sup>-1</sup> for the inverse, to be interpreted according to the operation in question

    With this in mind, we see that if a, b are in G, ab = c implies c is in G. We also see that ae = a , and aa<sup>-1</sup> = e. Note that these last equalities refer to what are called right identity and right inverse, respectively. The left identity and inverse follow follow from the group axioms. (Anybody want to try the simple proof?)

    Now, perhaps the most important distinction we can make between different sorts of groups is this: if the operation on the elements h, g of a group G is commutative (i.e. gh = hg) the group is said to be abelian, otherwise non-abelian. I referred earlier to a notational convention: for abelian groups the operation is said to be addition.

    Now non-commutivity need not throw us into a panic - think matrix multiplication for example, but the notation can be misleading at first. ab = c need not imply that ba = c, but it is always the case that ab = c means a = cb<sup>-1</sup> = b<sup>-1</sup>c, by my axioms above.

    One more word on commutivity; it, or the lack of it, is a property of the group elements, not of the operation (associativity would not follow otherwise).

    Roughly speaking, the order of a group G is the number of elements in the underlying set |G|. There's an interesting theorem of my buddy Lagrange, but first I need to tell you this.

    Consider the groups G and H. If every element of H is also an element of G, clearly G = H iff every element of G is also an element of H. But if there are elements in G not in H we may say H is a subgroup of G. As H is a group in its own right it must share the identity e with G, must have an inverse and must be closed under the group operation. Evidently the operation on G and H must be the same.

    OK. Lagrange says that the order of any subgroup H of G divides the order of G. This is less easy to prove than it looks, but it's a really cool result.

    I'm going to close with some examples of groups and things which aren't groups:

    Z, the integers, is an additive abelian group

    The even integers are an additive abelian group; the odds are not a group, neither are the primes

    S<sub>n</sub>, the set of permutations on n objects is a group, non-abelian for n > 2

    The set of rotations in 3-space are a non-abelian group.

    Tired of typing, more another time if anyone wants


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  3. #2  
    Forum Radioactive Isotope MagiMaster's Avatar
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    a = cb<sup>-1</sup> = b<sup>-1</sup>c
    I don't think this is right. ab = c doesn't imply a = b<sup>-1</sup>c in general. It does imply a = cb<sup>-1</sup> and b = a<sup>-1</sup>c though.


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  4. #3  
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    ditto, a<sup>-1</sup>b = ba<sup>-1</sup> implies commutativity.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  5. #4  
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    Woops, thanks for pointing that out MagiMaster and river_rat. What I said is true only in an abelian group (*blush*)
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  6. #5  
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    OK, if anyone is still awake, I have a lot more to say about subgroups, but first I think I need to do this.

    When thinking about group theory it is tempting to return to our favourite groups, R (the reals), N (the natural numbers), and Z (the integers). But this can be misleading in lots of things I now want say. For example, a lot of stuff which seems self-evident in R, Z and N actually may not apply to other groups. So, to give a taste of a quite different sort of group, I will introduce you to an important and versatile group.

    Consider the set {1, 2, 3}. Now consider the set {1, 3, 2}. It's not too hard to see what I've done, and let's write that exchange as (2, 3) i.e. exhange 2 with 3 and 3 with 2 (keep a close eye on the way brackets are used here!).

    That operation is referred to as a permutation, so let's try to figure out how many permutaions there might be on the set {1, 2, 3}.......pause for thought......That's right, for each set with n elements, there are n! permutations.

    What we have here, then, is a set with 3 elements whose permutations are themselves the elements of a 6 element group called the symmetric group S<sub>3</sub>. In fact basically this is matrix algebra. It is easy to see that this is a non-abelian group, as are all S<sub>n</sub> for n > 2. The connection with matrix algebra is really nice, which I'd be happy to expand upon (if asked).

    OK, to return to subgroups. If, for some group G with subgroup H we have that ghg<sup>-1</sup> is in H, for all g in G, all h in H, H is said to be a normal (sometimes called invariant) subgroup. Note that, as this must be true for all h in H, this condition is often written gHg<sup>-1</sup> = H. Note also this implies that gH = Hg, whether our group is abelian or not.

    Also note that every group G has at least one normal subgroup - G itself! Where the only normal sub-group of G is G, G is said to be a simple group.

    So here's another definition: gH and Hg are called the left (resp. right) cosets of H when g is not in H. This is an incredibly important construction, so let's see a simple example. Consider the group 3Z, those integers exactly divisible by 3. This is an abelian subgroup of Z, so has 0 as the identity e, addition as the operation. The left cosets of 3Z are 1 + 3Z, 2 + 3Z, 4 + 3Z, 5 + 3Z etc. Or are they? Elements of 3Z are ...-3, 0, 3, 6 etc. What's the difference between 1 + 6 and 4 + 3? This brings us to something really, really important.

    I invite you to look at these cosets. Notice anything? What's the relation between 1, 4, 7....on the one hand and 2, 5, 8....on the other? Yup, they differ by an element of 3Z. This defines what is called an equivalence relation. To formalize: given a normal subgroup H of a group G, gh<sup>-1</sup> is in H defines an equivalence relation(written ~) on G iff the relation is

    reflexive: g ~ g
    symmetric: g ~ h ==> h ~ g
    transitive: g ~ h and h ~ k ==> g ~ k

    for all g, h, k in G.

    This is an example of a universal construction, and is highly important. More later, if anybody wants.
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    Using the notation I gave earlier, if g ~ h ~ k, we may form an equivalence class {g, h, k}, for which we can elect a class representative, say g. Then we write that class as [g], it being understood that g, h and k are in [g].

    Note that, by the property of transitivity above, any element g of G can only be in one class: if g is in [g] and g is in [j], then [g] = [j]. This is thus a partition of G (strictly of the underlying set of G) into disjoint subsets. Here's the nifty bit.

    Ah no, wait, let's get concrete (but remember the caveat I gave when introducing the permutation groups). I talked about the normal subgroup 3Z of Z, that being the elements of Z exactly divisible by 3. We saw that 1 ~ 4 ~ 7 (for, 1 + 6 = 4 + 3 etc.) and 2 ~ 5 ~ 8, and we noted that not only did elements of 3Z differ by a factor of 3, so did the elements of the sets {...., 1, 4, 7...} and {...., 2, 5, 8....}.

    So we have 3 equivalence classes: [1], [2] and [3]. But wait! The subgroup 3Z has as elements {...-6, -3, 0, 3, 6,...} so let's run the election for class representative again: [0], [1] and [2], let's say. So if we gather these guys together we have an identity in [0]. We also know that any member of [1] plus any member of [2] is a member of [0] etc. so we have a closed operation too. Inverses are similarly easily checked.

    In other words, the collection of equivalence classes is itself a group which satisfies all the usual group axioms. Its called the quotient group. It is written G/~. That's the nifty bit.

    To sum up. For any group G, the set comprising a normal subgroup H of G together with its cosets forms a group, the quotient group G/~. This is an incredibly important construction, and not only in group theory.
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  8. #7  
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    Its funny to think that the existence of something as basic as a transversal for an equivalence class needs the axiom of choice
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  9. #8  
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    Oh hilarious. Anyway, to continue, where were we?

    Oh yes. To each group G we can associate its underlying set, which I will denote |G|. Now we know that functions on sets send elements of one set to elements of another: f: X → Y Now, given a function on |G|, we may have that f: |G| → |H|, where |H| is the underlying set of some group H. But groups carry a notion of structure, given by the axioms of the theory. So we require our function f to be rather more than a function on sets.

    So let's have f: G → H, where G and H are now groups. If f preserves the operational relationship between elements of G and H and is surjective, it is called a homomorphism. Specifically, let the operation on G be · and that on H be *, if we have f(g · g) = f(g) * f(g) for all g in G and all f(g) in H, f is a homomorphism. (Note that we do not require · and * to be different operations).

    Here's a familiar example. log(xy) = log(x) + log(y). This is in fact an example of a function that is surjective and also injective i. e. bijective. In these circumstances, we can assume f has an inverse, and f: G → H is called an isomorphism: a one-to-one, operation-preserving map.

    OK. One of the most important theorems in group theory follows.

    Remember I talked about the permutation (a.k.a the symmetric) groups?
    Some dude called Cayley came up with this:

    Every group of order n is isomorphic to a subgroup of the symmetric group S<sub>n</sub> (order n!)

    The proof is straightforward, but rather messy.

    Right, soon we need to see what it means for a homomorphism f to be "well defined", then we'll see if AC really is invoked in the partion into equivalence classes! But I have to run now, stay tuned.
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  10. #9  
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    Seeing that we have covered all the areas that you need to prove the quite useful fermat's little theorem i thought it would be useful to do that now.

    For those who are wondering, fermat's little theorem says that for any prime p and any integer a we have that a<sup>p-1</sup> is divisible by p (well not quite but it is close enough for a forum 8) )

    To start off we need the multiplicative group of integers modulo p (denoted Z<sub>p</sub>), which is the equivalence class of integers given by the equivalence relation a~b iff (a-b) is divisible by p with multiplication given by [a]*[b] = [a*b]. Some basic number theory shows that this is a group (for inverses i think you need that if GCD(p, q) = 1 then there exists integers f and g such that f*p + g*q = 1) and its identitity is 1.

    Now the order of the group Z<sub>p</sub> is p-1 as the set {1, 2, ..., p-1} is a transversal for the equivalence class. Now let a be an integer, then [a] is a member of of Z<sub>p</sub> and so the order of [a] divides the order of Z<sub>p</sub> for any prime p by Lagrange's Theorem.

    Expanding that gives the equation a<sup>k</sup> = 1 (mod p) where k is the order of the the element [a]. But k divides p-1 so a<sup>p-1</sup> = a<sup>k*t</sup> = (a<sup>k</sup>)<sup>t</sup> = 1<sup>t</sup> = 1 (mod p)
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  11. #10  
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    Quote Originally Posted by river_rat
    so a<sup>p-1</sup> = a<sup>k*t</sup> = (a<sup>k</sup>)<sup>t</sup> = 1<sup>t</sup> = 1 (mod p) :)
    Nice, but may I just note that, although this is the version of the little theorem I learned, it is not quite how you introduced it. Ah well, you did say.

    Anyhow, it reminds me to plug a gap which perhaps I should have done earlier, as the example I have been using to illustrate the quotient group uses the notion of a cyclic group implicitely. So, up front, let's say that a group is cyclic iff you can "go through" it and come back to the beginning. This isn't very clear, is it? OK, let's get dirty.

    It is a consequence of the axioms of our theory that any group G can be generated by repeated applications of the group operation to some subset |H| of the underlying set |G| of G. We can write this as |H|<sup>n</sup>, provided we remember that the operation may be addition.

    So, a cyclic group is one where the group generator is a single element, say g. We say that G is cyclic, order n, iff G = g<sup>n-1</sup>. The convention is that g<sup>0</sup> is defined to be the identity e. Note also that we require n to be integer (obviously) and that ∞ is integer. The obvious example of a cyclic group of infinite order is Z itself.

    From this definition, it should be clear that we will have g<sup>0</sup> = g<sup>n</sup> = e. We can now make contact with the quotient groups, but first I have to burden you with some notational conventions.

    Recall I wrote the generic quotient group as G/~. Now, this doesn't help a great deal unless we know what the ~ represents. Let's return to our old friend 3Z, integers exactly divisible by 3. We said that this subgroup of Z induces a partition of Z into equivalence classes. We can write this group more explicitely as Z/3Z, or more succinctly as Z<sub>3</sub>, they all mean the same thing.

    So let's look at Z/3Z. Elements are the equivalence classes [0], [1] and [2]. Let's start counting from -3 gazillion to 3 gazillion. Ready? [0], [1], [2], [0]. Done! Can you see what's going on? We'll take a peek inside the boxes in a bit, then all will be revealed. But it should clear from this that Z/3Z is a cyclic group order 3, but first some more notation.

    If I say 1 = 100 mod 99, what I really mean is 1 and 100 are the same providided I am willing to ignore the small matter of 99. So, what's in our boxes? [0] = {...,-6,-3, 0, 3, 6,...}, so any element of [0] = any other element of [0] mod an element of [0]. What about [1]?

    [1] = {...-5, -2, 1, 4, 7,...} so any element of [1] = any other element of [1] mod an element of [0]. Likewise for for [2]. This is the meaning of equivalence induced by a subgroup. Neat, huh?
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  12. #11  
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    So, there is lot more that could be said about group theory. But, before I proceed (if I do) let me just say this.

    I am amazed that there have been no questions about what we have written here. Yes, I well remember not wanting to appear foolish in front of my fellow-students, so I didn't ask either. Until I realized they felt exactly the same way!

    Anyway, I know of no-one who thought this subject easy on first encounter, so if you are trying to follow, don't worry if you have problems, we all did at first!

    Just ask. I will answer if I can, and others will, I'm sure, if I can't.
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  13. #12  
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    got unsure here and pretty soon i was lost.

    Quote Originally Posted by Guitarist
    OK, to return to subgroups. If, for some group G with subgroup H we have that ghg<sup>-1</sup> is in H, for all g in G, all h in H, H is said to be a normal (sometimes called invariant) subgroup.
    i can't think of a case where the subgroup is not a normal subgroup, can you provide an example?
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  14. #13  
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    Sure.

    Let's first reiterate the definition. A subgroup H of G is normal iff ghg<sup>-1</sup> is in H for all h in H, all g in G. This implies gH = Hg, so subgroups of an abelain group are normal because gh = hg for all h and g in G. We may use this as another definition of normalitty.

    So, to look for non-normal subgroups, we are better off looking at the subgroups of non-abelian groups. (In what follows you will for sure need pencil and paper, if you want to follow completely!)

    Consider the symmetric group S<sub>3</sub>, this being the 3! = 6 permutations of the 3-element set {1, 2, 3}. Elements of this group are witten as (321) etc, which simply means " replace 3 with 2, 2 with 1 and 1 with 3". (Note carefully the different styles of parenthesis).

    The group operation is simply two or more successive permutations, say (13)(12), but note two things. First, by convention, we work right-to-left, id est do (12) first then do (13) on the result.

    Second, it makes a big difference which order we do this: (12)(13) = {3, 1, 2} = (321) but (13)(12) = {2, 3, 1} = (123). So this is not an abelian group.

    By performing the group operation we see that (123)(321) = (321)(123) = e, (321)(321) = (123) and (123)(123) = (321), so we a have a subgroup {e, (123), (321)}. We also see that (12)(12) = e, so we may have {e, (12)} say as another subgroup.

    Note that as, say (13)(13) = e then (13) = (13)<sup>-1</sup>, we can find that (13)(12)(13)<sup>-1</sup> = (13)(12)(13) = {1, 3, 2} = (23) which is not an element of {e. (12)} (gHg<sup>-1</sup> not in H) so {e, (12) is not a normal subgroup.

    Alternatively, as (12)(13) = (321) yet (13)(12) = (123), gH notequal Hg, so {e, (12)} is not a normal subgroup.
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  15. #14  
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    much clearer, thank you.
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  16. #15  
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    No sweat. Anyway, having talked at excrutiating length about the non-abelian group S<sub>3</sub>, I was minded to talk about another famous non-abelian group, the group of rotations in n-space, n > 2. But these, while fascinating, are matrix groups, which would be hard to explain typograhically here.

    So let's do this instead.

    I talked about cyclic groups, and said that for g in G, if there is some equality such that g<sup>0</sup> = g<sup>n</sup> = e, then G is cyclic, order n. This last equality is what is called a group presentaion.

    Let's go back to our old friend Z/3Z. Here we have g<sup>3</sup> = g<sup>-3</sup> = e, this is our group presentation. Generally, one doesn't include the "= e" bit, it is understood.

    Now it's not rocket science to see that if g<sup>3</sup> = e and g<sup>-3</sup> = e, then g<sup>3</sup>g<sup>-3</sup> = ee = e. This is nice, it accords with our instinct (alas, not always reliable!), but it also fits with our definition of the inverse in group theory: if gh = hg = e then g and h are inverses mutually.

    So, let's invent our own rule: if an element g of a group is written right next to its inverse, then they cancel identically. You may think the highligted bit of our rule is pushing pedantry too far. But wait!

    We may also have a group with two or more generators (never mind for now what sort of group that might be - they do exist) So, for simplicity, let's use only unit indexes: gh = e, g<sup>-1</sup>h<sup>-1</sup> = e. Then we might hope that, as ee =e, we will have (gh)(gh)<sup>-1</sup> = ghg<sup>-1</sup>h<sup>-1</sup> = e. But, by our rule, g and its inverse cannot "see" each other, likewise h.

    So, we must have that (gh)<sup>-1</sup> = h<sup>-1</sup>g<sup>-1</sup>, whereby ghh<sup>-1</sup>g<sup>-1</sup> = e. This is called (by me only) cancelling from within.

    But - if it is the case that ghg<sup>-1</sup>h<sup>1</sup> = e, then I will have (gh)(hg)<sup>-1</sup> = e by the above, which, by the ordinary rules of arithmetic I may write as gh = hg, showing that this is the presentation of an abelian group.

    I think that's all pretty cool: the group presentation tells me the order of my group, whether it's cyclic and whether it's abelian. That's a lot of information.
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    Ah well.
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  18. #17  
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    Quote Originally Posted by Guitarist
    So here's another definition: gH and Hg are called the left (resp. right) cosets of H when g is not in H. This is an incredibly important construction, so let's see a simple example. Consider the group 3Z, those integers exactly divisible by 3. This is an abelian subgroup of Z, so has 0 as the identity e, addition as the operation. The left cosets of 3Z are 1 + 3Z, 2 + 3Z, 4 + 3Z, 5 + 3Z etc. Or are they? Elements of 3Z are ...-3, 0, 3, 6 etc. What's the difference between 1 + 6 and 4 + 3? This brings us to something really, really important.

    I invite you to look at these cosets. Notice anything? What's the relation between 1, 4, 7....on the one hand and 2, 5, 8....on the other? Yup, they differ by an element of 3Z. This defines what is called an equivalence relation. To formalize: given a normal subgroup H of a group G, gh<sup>-1</sup> is in H defines an equivalence relation(written ~) on G iff the relation is

    reflexive: g ~ g
    symmetric: g ~ h ==> h ~ g
    transitive: g ~ h and h ~ k ==> g ~ k

    for all g, h, k in G.

    This is an example of a universal construction, and is highly important. More later, if anybody wants.
    i'm confused and i've been looking at this for 2 hours now .

    for g+H = H+g where g is an element of the integers which are not an element of 3Z, and H is the set 3Z.
    what i first take note of is that g+H is an element of G, or another g. which makes it hard to get my head around gh<sup>-1</sup> being an element of H instead of G? to define the equivalence relation.
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    Hi Wallaby.

    I'm sorry to be the cause of your headache. In defining the equivalence relation gh<sup>-1</sup> is in H I had not intended to restrict h to the subgroup H. g and h are any element of G. I agree it's slightly ambiguous, but, in my defence, I do say at the end of the paragraph "for all g, h, k in G"

    Sorry about that, maybe I should have been more explicit.
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  20. #19  
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    Quote Originally Posted by Guitarist
    Hi Wallaby.

    I'm sorry to be the cause of your headache. In defining the equivalence relation gh<sup>-1</sup> is in H I had not intended to restrict h to the subgroup H. g and h are any element of G. I agree it's slightly ambiguous, but, in my defence, I do say at the end of the paragraph "for all g, h, k in G"

    Sorry about that, maybe I should have been more explicit.
    phew thats certainly relieving to hear.

    i'll have to read ahead in future instead of looking at the same thing for hours, silly me .
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    Quote Originally Posted by river_rat
    For those who are wondering, fermat's little theorem says that for any prime p and any integer a we have that a<sup>p-1</sup> is divisible by p (well not quite but it is close enough for a forum 8) )
    Not quite! If a is divisible by p, the result breaks down. You must include the condition that a and p are coprime for the theorm to work.

    Quote Originally Posted by river_rat
    Now the order of the group Z<sub>p</sub> is p-1 as the set {1, 2, ..., p-1} is a transversal for the equivalence class. Now let a be an integer, then [a] is a member of of Z<sub>p</sub> and so the order of [a] divides the order of Z<sub>p</sub> for any prime p by Lagrange's Theorem.
    Not really! If p divides a, then [a] is not a member of {[1],[2],...,[p-1]}.
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    Quote Originally Posted by river_rat
    For those who are wondering, fermat's little theorem says that for any prime p and any integer a we have that a<sup>p-1</sup> is divisible by p
    Incorrect! Fermats little theorem says that , not . Careless eh, river_rat?

    (And you must also have as pointed out. Example: .)
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  23. #22  
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    Quote Originally Posted by JaneBennet
    Fermats little theorem says that .
    If you multiply by a, you get . I think this is called Euler's theorem, or something. It has the advantage that a and p don't have to be coprime.
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    Hmm. I see that homomorphisms and normal subgroups, and even quotient groups, have been mentioned in this thread, but no mention of this

    Theorem: If G and H are groups and is a homomorphism, then the kernel of , , is a normal subgroup of G and the quotient group is isomorphic to a subgroup of H.

    Id have thought this was quite a useful result by itself. :?

    Anyway, proof: , . Hence .

    Define .

    Then , . Thus is well defined and injective.

    And . Hence is a monomorphism (injective homomorphism) and so is isomorphic to a subgroup of H. Namely the subgroup .
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  25. #24  
    . DrRocket's Avatar
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    Quote Originally Posted by JaneBennet
    Hmm. I see that homomorphisms and normal subgroups, and even quotient groups, have been mentioned in this thread, but no mention of this

    Theorem: If G and H are groups and is a homomorphism, then the kernel of , , is a normal subgroup of G and the quotient group is isomorphic to a subgroup of H.

    Id have thought this was quite a useful result by itself. :?
    Not only is that a useful result, it is probably the main reason that mathematicians care about normal subgroups. If it were not for that theorem, you probably would never had heard of normal subgroups.
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  26. #25  
    Forum Ph.D.
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    Yeah, that and the fact that every normal subgroup is the kernel of some homomorphism namely, if , N is the kernel of the homomorphism .
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