Re: > A group of order 15 is cyclic

k.hofmann wrote:

To prove: a group G of order 15 is cyclic.

The proof I am studying runs as follows:

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.

It's enough to show that Z(G) is nontrivial, for then G/Z(G) is cyclic,
hence G is abelian.

If Z(G) = 1, then G must contain 1 conjugacy class of size 5, and 3 of
size 4.

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

Assume first 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 and we're done.

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

From this, how do we arrive at a contradiction?

Do we really have 12 elements of order 5, namely z^j, w^j and y^j, as j
runs from 1 to 4? If so, how does this contradict the class equation
(whenever Z(G) is assumed to be trivial)?

I didn't really understand your argument.
What are x and y?
If they are general elements of the group,
why does y = y^{j^3}?

In answer to your last question, a cyclic group of order mn
contains exactly one subgroup of order m,
and this contains all the elements of order m;
so there are certainly not 12 elements of order 5 in your group.

Have you noted that if the group is not cyclic
then all elements have order 1, 3 or 5?
and if x,y are of order 5 then the subgroups they generate
are either the same, or else have only 1 in common.
So the number of elements of order 5 is a multiple of 4.
Similarly the number of elements of order 3 is even.
And the elements in each of your conjugacy classes
must have the same order.

Timothy Murphy
e-mail: gayleard /at/
tel: +353-86-2336090, +353-1-2842366
s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland

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