Prove: If $|G| = p^2$, then $G$ must be abelian.

Question

Exercise 15.G.5 from Pinter says:

Prove: If $|G| = p^2,$ $G$ must be abelian. (Use the preceding Exercise F.)

From earlier in the exercise, we have the following:

Let $G$ be a group whose order is a power of a prime $p,$ say $|G| = p^k.$ Let $C$ denote the center of $G.$

This exercise directs us to use "preceding Exercise F". Exercise 15.G.F states:

Conclude that if $G/C$ is cyclic, then $G$ is abelian.

So now our proof changes to a simpler case:

Prove: If $|G| = p^2,$ $G/C$ must be cyclic.

I.e.: if we can show that $G/C$ is cyclic, then we know that $G$ is abelian.

15.G.4 states:

$$|C|\textrm{ is a multiple of }p.\tag{1}$$

By Chapter 13, Theorem 3 (Lagrange's theorem):

$|G|$ is a multiple of $|C|.$

We know that:

$$|G| = p\cdot p.\tag{2}$$

From $(1),$ $(2),$ and Lagrange's theorem, we know that either $|C| = p$ or $|C| = p^2.$

If we consider the case where $|C| = p,$ then we know that

\begin{eqnarray}\textrm{Number of cosets of }C\textrm{ in }G &=& \textrm{Index of }C\textrm{ in }G\\ &=& (G:C)\\ &=& |G|/|C|\\ &=& p^2/p\\ &=& p.\end{eqnarray}

$G/C$ is the set of all the cosets of $C$ in $G,$ so that means $$|G/C| = p.$$

By Chapter 13, Theorem 4:

If $G$ is a group with a prime number $p$ of elements, then $G$ is a cyclic group.

We therefore know that $G/C$ is a cyclic group.

At this point, it seems we have taken the path suggested by the book, i.e.: $$|G| = p^2\:\implies\:G/C\textrm{ is a cyclic group}\:\implies\:G\textrm{ is abelian}.$$

However, I see one issue here.

In order for $G$ to be abelian, all $x \in G$ must also be in $C.$ I.e." $|C| = |G|,$ and thus $C = G.$ However, in the above steps, $|C| = p$ is obviously smaller than $|G| = p^2.$

So, I must be missing something here. :-) Suggestions welcome!

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Note: there are a few other questions on this site which deal with similar exercises. However, what I'm asking here is different from those for the following reasons:

• The approach in those questions doesn't quite match the approach suggested in Pinter
• Those questions don't specifically deal with my question about why $|C| \neq |G|.$

Answer 1

You have appropriately followed Pinter's suggestion. Whether $|C|=p$ or $|C|=|G|,$ we would be able to conclude that $G/C$ is cyclic--in the former case, because it has prime order; in the latter, because it is the trivial group. Thus, $G$ is abelian, and so it is the case that $C=G.$

Another way to look at it is this: the $|C|=p$ case was merely a possibility that had to be considered, which turned out to be impossible, since $$|C|=p\:\implies\:|G/C|=p\:\implies\:G/C\textrm{ is cyclic }\:\implies\:G\textrm{ is abelian}\:\implies\:|C|=|G|=p^2.$$ In other words, the case $|C|=p$ led to a contradiction, so we're left with the case $|C|=|G|,$ meaning that $G$ is abelian.