I have a 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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    $\begingroup$ One can prove that $p$-groups have non-trivial centres using the conjugacy class equation. Is this unacceptable, by the last line? $\endgroup$ Sep 14 '11 at 3:48
  • 3
    $\begingroup$ 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. $\endgroup$ Sep 14 '11 at 3:54
  • $\begingroup$ @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. $\endgroup$ Sep 14 '11 at 5:08
  • $\begingroup$ @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. $\endgroup$ Sep 14 '11 at 5:55
  • $\begingroup$ In his question, he examines the case where |Z(G)| = p. However, in that case, |Z(G)| < |G|, which means that not all items in G are in the center, which of course means that G is not abelian. So, why would he consider the case of |Z(G)| = p? $\endgroup$
    – dharmatech
    Jul 17 '19 at 4:38

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.

  • $\begingroup$ 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). $\endgroup$ May 26 '14 at 8:13

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 $G$. $\Box$

  • $\begingroup$ @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. $\endgroup$ Sep 14 '11 at 5:08
  • $\begingroup$ 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? $\endgroup$ Sep 15 '11 at 3:06
  • $\begingroup$ 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 $a$. $\endgroup$
    – Deven Ware
    Sep 15 '11 at 3:48
  • $\begingroup$ @Jimmy Valmer: I have just re-edited again as I realize I may have misinterpreted your question the first time I updated my post. $\endgroup$
    – Deven Ware
    Sep 15 '11 at 6:38

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.

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

First we need the following lemma:

If $G/Z(G)$ is cyclic then G is abelian.

You can check the prove in here: https://yutsumura.com/if-the-quotient-by-the-center-is-cyclic-then-the-group-is-abelian/.

Then you need this theorem:

If $p$ is a prime and $P$ is a group of prime power order $p^{\alpha}$ for some $\alpha \geq 1$, then $P$ has a nontrivial center: $Z(P) \neq 1$.

You can check the prove in the answer above: https://math.stackexchange.com/a/64374/435467.

Now is the prove of the question:

Since $Z(P) \neq 1$, and as the order of $P$ is $p^2$, there's no other choice that $|Z(P)| = p$. Then $|P/Z(P)| = \frac{p^2}{p} = p$, and as any group with prime order must be cyclic, $P/Z(P)$ is cyclic. By the above lemma, $G$ must be abelian.

  • $\begingroup$ why can't $|Z(P)|=p^2$? $\endgroup$ Aug 26 at 18:26

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.

  • $\begingroup$ I have posted an answer below , could you please check it ? $\endgroup$
    – user228168
    May 20 '16 at 6:38

I will present a different approach which only involves group actions.

If there is an element of order $p^2$ then the group is the cyclic group of order $p^2$ which is, of course, abelian.

Otherwise, suppose every element in $G$ has order less than $p^2$, it is equivalent to say that the order of every non-unit element in $G$ is $p$. Now we choose an arbitrary non-unit element $g\in G$, then we have $|\langle g \rangle|=p$. Suppose $\Omega=\{a_1\langle g \rangle,a_2\langle g \rangle,...,a_p\langle g \rangle\}$ is a left cosets partition of $G$. Let $\langle g \rangle$ acts on $\Omega$ by the natural left multiplication. Hence $$ |\text{Orbit}(a_i\langle g \rangle)|=\frac{|\langle g \rangle|}{|\text{Stab}(a_i\langle g \rangle)|}=\begin{cases}p,&\text{when } |\text{Stab}(a_i\langle g \rangle)|=1\\ 1,&\text{when }|\text{Stab}(a_i\langle g \rangle)|=p\end{cases} $$ But $|\text{Orbit}(\langle g \rangle)|=1$, so for every left coset $a_i\langle g \rangle$, $|\text{Orbit}(a_i\langle g \rangle)|=1$ which means for every $g^k,k\in\mathbb Z$ and $a_i$, $a_i^{-1}g^ka_i\in\langle g \rangle$. Thus for every $a_ig^m$, $(a_ig^m)^{-1}g^k(a_ig^m)\in\langle g \rangle$ which implies $\langle g \rangle$ is a normal subgroup of $G$. Since $g$ is an arbitrary non-unit element in $G$, it follows that every subgroup of order $p$ of $G$ is a normal subgroup.

Now take $g_1,g_2\ne e$ where $e$ is the unit and $\langle g_1 \rangle\cap\langle g_2 \rangle=\{e\}$. From what we have proved above, we know that $\langle g_1 \rangle$ and $\langle g_2 \rangle$ are normal subgroups. Hence \begin{align} g_1g_2g_1^{-1}&=g_2^{\ell_1}\tag 1\\ g_2g_1g_2^{-1}&=g_1^{\ell_2}\tag 2 \end{align} for some $\ell_1,\ell_2\in\mathbb Z$. Note that $g_2g_1g_2^{-1}g_1^{-1}=g_1^{\ell_2-1}$ by $(2)$ but $g_2g_1g_2^{-1}g_1^{-1}=g_2^{1-\ell_1}$ by $(1)$. So $g_1^{\ell_2-1}=g_2^{1-\ell_1}\in\langle g_1 \rangle\cap\langle g_2 \rangle=\{e\}$. Therefore $g_1^{\ell_2}=g_1$ and $g_2^{\ell_1}=g_2$ and $g_1g_2g_1^{-1}=g_2$ for all $g_1,g_2\ne e$. Hence $G$ is abelian.


A slightly different way using actions and orbits.

Let $|G| = p^2$. Show that $G$ is abelian.

If there exists an element $g \in G$ so that $|g| = p^2$, then $g$ generates $G$, making $G$ cyclic and thus abelian, and we are done.

So, assume there is no element of $p^2$ power. Since the order of each element must divide $|G|$, all elements (except the identity) must be of $p$ power.

Let $a \in G$ be arbitrary. We want to show that $a$ commutes with every other element of $G$.
If $a$ is the identity, we are done, so assume otherwise. Then $|a| = p$.

Define $H = \langle a \rangle$, so $|H|=p$. Since $|G:H|=p$, and $G$ is a $p$-group, $H$ is normal in $G$. So, let $G$ act on $H$ by conjugation, and let $H_G$ denote the union of the single-element orbits of this action.

Since orbits partition, we now have $$ |H| = |H_G| + \sum_{|\mathcal{O}| > 1} |\mathcal{O}|, $$ where the sum on the right counts all elements of $H$ in non-singleton orbits (similar to the class equation, but we are only acting on the subgroup $H$).

Since the size of orbits must divide the size of the group doing the acting (in this case $G$), we can infer that every non-singleton orbit has size $p$ (since $p^2$ would be too big).

Looking at the union of singleton orbits, we see $H_G = \left\lbrace h \in H : \forall g \in G : g^{-1}hg = h\right\rbrace$. In other words, $H_G$ contains all elements $h$ that commute with every element of $G$.

Since the identity is in $H$, and the identity commutes with everything, $|H_G| \geq 1$, but since the smallest a "multi-element" orbit can be is $p$, there are no elements in multi-element orbits, and we have that $|H_G| = |H| = p$, implying that every element of $H$ (in particular, its generator $a$) commutes with every element of $G$.

Since $a$ was arbitrary, every element of $G$ must commute with every other element of $G$, and we are done.


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