# Showing non-cyclic group with $p^2$ elements is Abelian

I have a non-cyclic group $G$ with $p^2$ elements, where $p$ is a prime number. Some (potentially) useful preliminary information I have is that there are exactly $p+1$ subgroups with $p$ elements, and with that I was able to show $G$ has a normal subgroup $N$ with $p$ elements.

My problem is showing that $G$ is abelian, and I would be glad if someone could show me how.

I had two potential approaches in mind and I would prefer if one of these were used (especially the second one).

First: The center $Z(G)$ is a normal subgroup of $G$ so by Langrange's theorem, if $Z(G)$ has anything other than the identity, it's size is either $p$ or $p^2$. If $p^2$ then $Z(G)=G$ and we are done. If $Z(G)=p$ then the quotient group of $G$ factored out by $Z(G)$ has $p$ elements, so it is cylic and I can prove from there that this implies $G$ is abelian. So can we show theres something other than the identity in the center of $G$?

Second: I list out the elements of some other subgroup $H$ with $p$ elements such that the intersection of $H$ and $N$ is only the identity (if any more, due to prime order the intersected elements would generate the entire subgroups). Let $N$ be generated by $a$ and $H$ be generated by $b$. We can show $NK= G$, i.e every element in G can be wrriten like $a^k b^l$. So for this method, we just need to show $ab=ba$ (remember, these are not general elements in the set, but the generators of $N$ and $H$).

Do any of these methods seem viable? I understand one can give very strong theorems using Sylow theorems and related facts, but I am looking for an elementary solution (no Sylow theorems, facts about p-groups, centrailzers) but definitions of centres and normalizers is fine.

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One can prove that $p$-groups have non-trivial centres using the conjugacy class equation‌​. Is this unacceptable, by the last line? – Dylan Moreland Sep 14 '11 at 3:48
Hint: prove that the center of a non-trivial $p$-group is non-trivial and prove that if $G$ is a finite group such that $G/\textbf{Z}(G)$ is cyclic, then $G$ is abelian. – Amitesh Datta Sep 14 '11 at 3:54
@Dylan - I know the conjugacy class equation but I am looking for a simpler "clever" answer that maybe only works for non cyclic $p^2$ groups. – Jimmy Valmer Sep 14 '11 at 5:08
@Jimmy: I've merged your duplicate account into your original one, and implemented the comments you wanted to make. If you register your account, that should help prevent future login difficulties. – Zev Chonoles Sep 14 '11 at 5:55

Your first approach is good; The center of a $p$-group is non-trivial:

Proof: The center of any group is the union of the 1-element conjugacy classes in the group. For a $p-$group, the size of every conjugacy class is a power of p because the order of a conjugacy class must divide the order of the group. Then let $p^{n_i}$ be the order of the conjugacy classes, and the conjugacy class equation tells us that $|G| = p^n = |Z(G)| + \sum_i (p^{k_i})$, where $0 < k_i < n$ and thus $p$ must divide $|Z(G)|$ implying that the center is non-trivial. $\Box$

EDIT: Explaining class equation:

The conjugacy classes partition the group, so we know that $|G| = \sum|cl(a)|$. But as I said in the proof above, the union of the singleton conjugacy classes is $Z(G)$, so we can rewrite this equality as

$|G| = |Z(G)| + \sum|cl(a)|$

where we assume that each conjugacy class is represented only once (i.e we are not including the singletons in the second summand). However, since we know that the order of a conjugacy class divides the order of a group we can rewrite the second summand to be $\sum_i(p^{k_i})$ where $0 < k_i < n$ because clearly none of them can have the same order as $G$, and we finally get

$|G| = |Z(G)| + \sum_i(p^{k_i})$ where $0 < k_i < n$

Now to see that $p$ must divide $|Z(G)|$ we see that we can move the second summand to left and replace $|G|$ with $p^n$ to get $p^n - \sum_i(p^{k_i}) = |Z(G)|$ and clearly we can factor out $p$.

EDIT: Showing that the order of a conjugacy class must divide the order of a group:

To prove this, we will show that the size of a conjugacy class of a, $cl(a)$ is the index of the centralizer of $a$, $C(a)$.

Suppose $x$ and $y$ both make the same conjugate of $a$, or $xax^{-1} = yay^{-1}$. Then multiplying on the left by $y^{-1}$ and on the right by $x$ we can see that $y^{-1}xa = ay^{-1}x$ and hence, $y^{-1}x \in C(a)$. Thus we can also see that $x\in yC(a)$ and hence $xC(a) = yC(a)$.

Similarly, suppose $xC(a) = yC(a)$ then it follows that $x \in yC(a)$ and hence $x = yz$ for some $z\in C(a)$. Thus $xax^{-1} = (yz)a(yz)^{-1} = yzaz^{-1}y^{-1}$. But we know that $z\in C(a)$ so we can rewrite this as $yazz^{-1}y^{-1}$, or $yay^{-1}$. Thus $x$ and $y$ make the same conjugate of $a$.

Thus we have shown that the number of cosets of $C(a)$ equals the number of elements in $cl(a)$, or

$(G : C(a)) = |cl(a)|$

Since $C(a)$ is a subgroup of $G$, clearly $(G : C(a))$ divides $|G|$ and hence, $|cl(a)|$ divides $G$. $\Box$

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@Deven - That is the approach I had in mind incase I can't find the more "elementary" solution, but I am confident that the 2nd approach I listed is potentially fruitful. I have recieved previous questions of this nature using that method. For example, I can show groups of order 9 must be abelian through a method similar to the second method, but that requires me to do some specific case work that I can't seems to generalize for this case. – Jimmy Valmer Sep 14 '11 at 5:08
This answer seems the most elementary reformulation of the class equation approach, so I voted this as the accepted answer, but could you please elaborate on something for me? I don't see why the order of conjugacy classes in p-groups must divide the order of the group. Is this simple? – Jimmy Valmer Sep 15 '11 at 3:06
I have edited my post, is it more clear now that I write the class equation in this situation as $p^{n} = |Z(G)| + \sum_i(p^{k_i})$ where $0 < k_i < n$ ? if not you can also consider the class equation in an equivalent form here $p^{n} = |Z(G)| + \sum\frac{|G|}{|cl(a)|}$ where $cl(a)$ represents the conjugacy class of $a$. – Deven Ware Sep 15 '11 at 3:48
@Jimmy Valmer: I have just re-edited again as I realize I may have misinterpreted your question the first time I updated my post. – Deven Ware Sep 15 '11 at 6:38

Here is a way to show that the center of a group of order $p^2$ cannot be trivial without using the class equation. I think that the major drawback (and it is major) is that it is very specific for groups of order $p^2$. The class equation is much better because it is a much more general result

If $G$ has elements of order $p^2$, then the result follows because $G$ is cyclic. So suppose that all nontrivial elements of $G$ have order $p$.

Let $x\in G$ be of order $p$. Now, $\langle x\rangle$ is normal in $G$, since its index is the smallest prime that divides the order of $G$. Therefore, $yxy^{-1}\in\langle x\rangle$, so $yxy^{-1}=x^r$ for some $r$, $1\leq r \leq p-1$.

It is now easy to verify that $y^ixy^{-i} = x^{r^i}$. In particular, $y^{p-1}xy^{1-p} = x^{r^{p-1}}$. By Fermat's Little Theorem, $r^{p-1}\equiv 1 \pmod{p}$, so $y^{p-1}xy^{1-p} = x$. That is, $y^{p-1}$ centralizes $x$. But $y$ is of order $p$, so $y^{p-1}=y^{-1}$. Since $y^{-1}$ centralizes $x$, so does $y$. That is, $yx=xy$. Thus, the centralizer of $x$ contains at least $\langle x\rangle$ and $y$, hence is of order at least $p+1$. Since its order must divide $p^2$, the centralizer of $x$ is all of $G$, so $x$ is central.

Thus, $Z(G)$ is nontrivial.

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Neat way to do it. An alternative way to get the final conclusion is to take some element not in the chosen subgroup and repeat the argument, giving two normal subgroups that intersect trivially, which means that they centralize each other (and since they generate the entire group, this makes them central). – Tobias Kildetoft May 26 '14 at 8:13

So Deven's helpful comment shows why the center of any $p$-group is nontrivial, so look back at your first approach.
Hint: if $Z(G) = p$, consider an element $x \in G$ that is not in $Z(G)$.
What is the overlap between $Z(G)$ and the cyclic subgroup of $G$ generated by $p$? What must the centralizer of $x$ be? What does this imply about $Z(G)$?

Edit: This is an alternative to the path of showing that $G/Z(G)$ cyclic implies $G$ abelian; I consider this route to be also as illuminating and even elegant.

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JakeR - I already know how to handle the problem if $Z(G)=p$ or $p^2$, so I really need a reason why $Z(G)>1$. – Jimmy Valmer Sep 14 '11 at 5:08

Here is an alternative way to show this which does not show that the center of a finite $p$-group is non-trivial. The proof is far from elementary, though the results I will use are of more general use.

In fact, what I will show is that if $G$ is a finite $p$-group then $|G/G'|\equiv |G|\pmod {p^2}$ (so a finite group of order $p^n$ is solvable of derived length at most $\frac{n}{2}$ rounded up).

To do this we will be be using some results about complex characters of finite groups (so this answer will hopefully serve as a (slight) motivation to learn more about this topic). I will not go through the definitions of irreducible complex characters here, but if anyone wants to see them, they can start at http://en.wikipedia.org/wiki/Character_theory and work their way through the various parts involved.

The results we need for this are:

If $Irr(G)$ denotes the set of irreducible complex characters of $G$ then $$|G| = \sum_{\chi\in Irr(G)}\chi(1)^2$$

If $\chi\in Irr(G)$ then $\chi(1)$ divides $|G|$.

And finally that $|G/G'| = |\{\chi\in Irr(G)\mid \chi(1) = 1\}|$.

Applying this to a finite $p$-group, we see that if $\chi\in Irr(G)$ with $\chi(1)\neq 1$ then $p$ divides $\chi(1)$, and we get $$|G| = \sum_{\chi\in Irr(G)}\chi(1)^2 = \sum_{\chi\in Irr(G),\, \chi(1) = 1}\chi(1)^2 + \sum_{\chi\in Irr(G),\, \chi(1)\neq 1}\chi(1)^2$$ $$= |\{\chi\in Irr(G)\mid \chi(1) = 1\}| + p^2m = |G/G'| + p^2m$$ for some natural number $m$, which proves the original claim.

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