Abelian Groups CyclicDate: 03/05/2002 at 11:48:24 From: kristin bedell Subject: Abelian Groups Prove that every abelian group of order 6 is cyclic. I know the non-abelian group of order 6, but I don't know how to find it. Help please - I've been racking my brains trying to figure this out. -Kristin Date: 03/05/2002 at 18:13:55 From: Doctor Paul Subject: Re: Abelian Groups Suppose by way of contradiction that |G| = 6 and G is not cyclic. Then G has no elements of order six. The identity is the only element of order one. By Lagrange's Theorem, the order of each element (which is the order of the subgroup it generates) must divide the order of the group so if |G| = 6 G is not cyclic then g in G implies |g| = 2 or |g| = 3. Claim: It cannot be the case that every non-identity element has order two. Proof: Assume that all non-identity elements have order two. Pick two such elements h and k. So h and k are distinct and both have order two. Notice that h^2 = 1 ==> h = h^(-1). Similarly, k = k^(-1) Let H = {1, h, k} be a subset (notice, not a subgroup since hk is not in H) of G. Notice that hk is not equal to h and hk is not equal to k: If hk = h then k = 1 (multiply both sides by h^(-1) on the left) which is not the case, and if hk = k then h = 1 (multiply both sides by k^(-1) on the right), which is not the case either. Also notice that hk is not 1, for if hk = 1 then h = k^(-1) = k, which is not true since we chose h and k to be distinct. So hk is not 1, hk is not h, and hk is not k. Notice also that hk has order two: hk is not 1 and (hk)^2 = h^2 * k^2 since G is abelian and this is clearly the identity. Thus hk has order two and is hence an element of G. So now let H' = {1, h, k, hk} be a subset of G. H' contains four distinct elements of G. But we know |G| = 6 so there exists m in G with m not in H' and |m| = 2 (G has six elements and we have only discussed four of them, so we must be able to pick another one). So m is not 1, m is not h, m is not k, and m is not hk. Also: mh and mk are going to be elements of G since G is closed under the group operation (mhk will also be in G, but we don't need this information). If we can show that mh and mk are distinct and not equal to m, and that they are not any of the elements in H', then we will have that G must contain *at least* seven elements: {1, h, k, hk, m, mh, mk} It may also be the case that mhk will be in G, but we would need to check to see if mhk is one of the other elements we already have. This isn't necessary to get the desired contradiction. Clearly if G contains at least seven elements, then we have a contradiction since we assumed |G| = 6. So let's show that mh and mk are distinct and not equal to m, and that they are not any of the elements in H'. First: mh and mk are distinct. For if mh = mk then h = k, a contradiction. Certainly mh is not m, because if this were the case, then h would equal 1, which we know isn't true. mh is not 1 because if mh = 1 then we have m = h^(-1) = h, a contradiction. mh is not h because if mh = h, then m = 1, a contradiction. mh is not k, because if mh = k, then m = k*h^(-1) = k*h = h*k, a contradiction. Finally, mh is not hk because if mh = hk, then hmh = hhk, which implies m*h^2 = h^2*k, which implies m = k, a contradiction. A similar argument shows that mk is not m, and that mk is not any of the elements in H'. So if every non-identity element of G has order 2, then we can get that |G| > 6, a contradiction. So it must the case that G cannot have all non-identity elements of order two. Now we want to argue that no non-identity element can have order three. Once we have this, then it must be the case that G contains at least one element of order two and at least one element of order three (since we will have shown that G cannot have all non-identity elements of order two and G cannot have all non-identity elements of order three and the only possibilities for the orders of non-identity elements in G are two and three). Let x and y be in G and let |x| = 2 and |y| = 3 (notice that this forces x^(-1) = x and y^(-1) = y^2). Then xy is in G (since G is a group) and we want to claim that |xy| = 6. Notice that: xy is not the identity, for if it were, then xy = 1, which implies x = y^(-1) = y^2. So we have x = y^2. Notice that |x| = 2 implies x = x^(-1). Substituting, we obtain: x^(-1) = y^2, which implies x*y^2 = 1, and we'll see below that this cannot be. (xy)^2 = x^2 * y^2 = y^2, which is not the identity; if y^2 = 1, then y = y^(-1), and we know that in this case y^(-1) = y^2, so we would have y = y^2, which would imply y = 1 (multiply both sides by y^(-1), which is plainly not true. (xy)^3 = x^3 * y^3 = x, which is not the identity. (xy)^4 = x^4 * y^4 = y, which is not the identity. (xy)^5 = x^5 * y^5 = x*y^2, which is not the identity; if x*y^2 = 1, then we can write: x = y^(-2) = [y^(-1)]^2 = [y^2]^2 = y^4 = y Plainly, x is not y because |x|=2 and |y| = 3. Finally, (xy)^6 = x^6 * y^6 = 1 * 1 = 1. So |xy| = 6. This is a contradiction since we assumed that G was not cyclic. Thus our assumption that G was not cyclic must be false, and it is in fact the case that G is cyclic. Now we come back to the unresolved issue above of showing that no non- identity element can have order three. Suppose by contradiction that every non-identity element has order three. Let h be in G and let H = {1, h, h^2} be a subset of G. Then the elements in H are distinct. Moreover, |G| = 6, so there exists k in G, k is not in H, and |k| = 3. Plainly, G contains k^2, and I claim that k^2 is not in H. If k^2 = 1, then k = k^(-1), which is not true since we know that k^(-1) = k^2 and k^2 is not k since |k| = 3. If k^2 = h, then square both sides to obtain: k = h^2, a contradiction of the fact that k is not in H. If k^2 = h^2, then square both sides to obtain: k = h, a contradiction of the fact that k is not in H. Thus we have H' = {1, h, h^2, k, k^2} is a subset of G. Again, |G| = 6 so there exists m in G, m is not in H', and |m| = 3. Similar arguments show that m^2 is not 1, m^2 is not h, m^2 is not h^2, m^2 is not k, m^2 is not k^2. Thus we have the subset of elements: {1, h, h^2, k, k^2, m, m^2} all in G. This cannot be in a group of order six. Notice that we have yet to consider any of the products of the form k*h, k*h^2, etc. But we didn't need to in order to show that G must contain more than 6 elements. Thus no non-identity element can have order three. This completes the (somewhat lengthy) proof. I hope it's clear. Please write back if you'd like to talk about this some more. - Doctor Paul, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/