# Re: -- Showing that a group of order 15 is abelian using brute force methods

• From: "Achava Nakhash, the Loving Snake" <achava@xxxxxxxxxxx>
• Date: Thu, 2 Oct 2008 22:46:15 -0700 (PDT)

On Oct 2, 10:07 pm, "k.hofmann" <boqui...@xxxxxxxxxxxxxxxx> wrote:
To prove: a group G of order 15 is cyclic.

The proof  is by 'brute force', following

http://www.math.rice.edu/~hassett/teaching/356spring04/solution.pdf

Since G contains an element x of order 3 and an element y
of order 5, the map Z/3 x Z/5 --> G defined by (i, j) --> x^i y^j is an isomorphism, provided xy = yx,
for then Z_3 x Z_5 ~  Z_{15}.

This much is clear.

What I gathered from the following paragraph is that it would be enough to show that Z(G) is nontrivial, for then G/Z(G)  would be cyclic, hence G is abelian.

If Z(G) = 1, then G must contain 1 conjugacy class of size 5, and 3 of size 3 (since there is only one class of size 1).

Set z = xyx^{-1} and w = x^{-1} y x.

Assume z =/= y^j for any j; then z generates a subgroup distinct from <x>, as does <w>. Furthermore, <z> and <w> are also distinct.

So we see that we have (at least) 12 elements of order 5, namely z^j, w^j and y^j, as j runs from 1 to 4?

How does this contradict the observation about the possible sizes of the conjugacy classes when we assume Z(G)  = 1?

On the other hand,  if  z = y^j,  then y = y^{j^3} so that j^3 = 1 mod 5 ==> j = 1 (mod 5) so that z = y ==> xy = yx.

This is what we wanted, but aren't we supposed to arrive at a contradiction at this point  as well?

I'm also a bit confused about the structure of the proof.

I hope someone can kindly clarify what's really going  on here.  Thank you in advance.

Will you never leave me alone! Now I have to solve your problem in an
elementary way. All right, all right, I will do it. You have shown
that there at least 12 elements of order 5, so there can only be 2 of
order 3. That means that the subgroup of order 3 (generated by x) is
normal, and so if y has order 5, then yxy^-1 = x, in which cases x
commutes with y, or yxy^-1 = x^2. In this cases (y^2)x(y^-2) = x^4,
(y^3)x(y^-3) = (x^4)^2 = x^8, ..., (y^5)x(y^-5) = x^32. But y^5 = 1,
so x = x^32, or x^31 = 1. But x^3 = 1, so x = 1, a contradiction. In
other words, x must commute with every element of order 5 and so is in
the center. This finishes your proof.

It is not clear to me how much you do or don't know, but this problem
becomes very easy if you happen to know Sylow's theorem. Then you
ssee that the subgroup of order 5 is normal, since all subgroups of
order 5, and the number of them must be congruent to 1 mod 5. It also
needs to be a divisor of the order of the group. That means there can
only be 1 of them and so it is normal. As for subgroups of order 3,
there can be 1, 4, 6, 10, 13, ... of them and yet also a divisor of
15, and so the subgroup of order 3 must also be normal. Then letting
x generate the unique subgroup of order 3 and y the unique subgroup of
order 5, by normality, xy(x^-1)(y^-1) in both the subgroup of order 3
and the subgroup of order 5 which implies that it is the identity.
Hence xy = yx and the group must be abelian, hence cyclic.

Hope this helps,
Achava
.

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