Why does an element of order p in G/<x> imply an element of order p² in G?

[Mathdope]

[SOLVED] General question on cosets

Assume a group G is finite, abelian. Let p^n be a divisor of |G|, p prime, n >1. Let x in G have ord(x) = p (by Cauchy's theorem).

Here's my question. If you now take the factor group of G/<x> you have an element of that factor group that is also of order p. Since the elements of the factor group are cosets of <x> in G, any element has the form <x>g, where g is in G. Thus it would seem that (<x>g)^p = <x>g^p = <x> for some g in G.

What additional fact(s) would make g have order p^2? Is the fact that <x>g has order p in G/<x> enough to guarantee that g has order p^2 in G?

What prevents g^p = e in G?

In particular, how do I know that somewhere in G lies an element of order p^2?

I expect to feel very foolish when I see the answer.

 

[StatusX]

First of all, it won't hold in general that G has elements of order p^2. For example, take $(\mathbb{Z}/p\mathbb{Z})^n$, which has order p^n, and all of whose elements have order p (or 1).

As far as the condition for this to happen, you summarized it when you said "What prevents g^p = e in G?" We can see g will have order p^2 iff this does not happen. One direction is trivial, and the other follows from the fact that g^p lies in <x>, so itself has order p, meaning g has order p^2.

In other words, if there is a single element in G with order kp, k>1, then there is an element with order p^2. You should be able to generalize thisargument to higher powers of p.

 

[Mathdope]

First of all, it won't hold in general that G has elements of order p^2. For example, take $(\mathbb{Z}/p\mathbb{Z})^n$, which has order p^n, and all of whose elements have order p (or 1).

Thank you. That's what I thought but it seemed as though the problem implied otherwise. This clears it up for me, as I'm no longer trying to prove something that doesn't hold in general (which is apparently why I couldn't get it to work!).