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

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!

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 $$|C| \neq |G|.$$

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.
• You're welcome! To be fair, I'm not sure I'd ever been consciously aware of the fact that the trivial group is cyclic before, but I may simply have let that memory lapse as unimportant. Since the trivial group so uninteresting, I haven't spent much time considering it at all! It's worth noting that Exercise 15.G.F is equivalent to stating: "If $G/C$ is cyclic, then it is trivial." Commented Jul 17, 2019 at 12:59