# If $G$ is a non-abelian group of order 27, what is the order of its commutator subgroup $G'$?

Maybe this question has been posted. Point out if so. In the book Character Theory Of Finite groups by I.Martin.Isaacs, the following is found at the page:25.

For instance, if $G$ is a nonabelian group of order 27, then $|G:G'|=9$ and $G$ has exactly 11 conjugacy classes.

Could someone explain to me? Thanks.

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I mean: I can understand nothing about that statement... Thanks for paying attention. – awllower May 23 '12 at 3:12
I have tried to argue that G' cannot be of order =9, but I understand not why it cannot be =G. Also, I figure nothing out about its conjugacy classes. – awllower May 23 '12 at 3:15

The commutator of a $p$-group cannot be the whole group. The center of a group of order $27$ is nontrivial, so $G/Z(G)$ must be abelian (since it has order either $1$ or $9$; it cannot be cyclic and nontrivial). Therefore, $[G,G]$ cannot be the whole group, since $G$ has a nontrivial abelian quotient.

(More generally, a $p$-group of order $p^n$ always has subgroups of order $p^{n-1}$ which must be normal; the quotient is abelian of order $p$, so the commutator subgroup must be contained in all subgroups of order $p^{n-1}$. In fact, the order of the commutator subgroup of a group of order $p^n$, $n\geq 2$, is at most $p^{n-2}$.)

So the commutator subgroup of a nonabelian group of order $27$ must be of order $3$ or $9$ (it cannot be of order $1$, since $G$ is assumed to be nonabelian).

To see that it cannot be of order $9$, note that there is certainly a normal subgroup of $G$ of order $3$: the center (center cannot be the whole group, since $G$ is nonabelian). So $G/Z(G)$ has order $9$, and hence is necessarily abelian, so the commutator subgroup must be contained in the center. Hence, the commutator subgroup is of order exactly $3$, so $[G:G'] = 9$.

To see that $G$ has exactly $11$ conjugacy classes, remember that the sum of of the orders of the conjugacy classes equals $|G|=27$. The conjugacy classes must have order a power of $3$. A conjugacy class has exactly $1$ element if and only if it corresponds to a central element, so you have $3$ classes of order $1$. The size of the conjugacy class is the index of the centralizer of the element; every element is centralized by elements of the center and by itself, so a noncentral element is centralized by at least $4$ elements, hence the centralizer has order exactly $9$ (cannot be $27$ because it's not central, must be a power of $3$, and cannot be $3$); thus, a noncentral element has exactly $3$ conjugates (the index of the centralizer).

So we have $3$ classes of order $1$ each, accounting for $3$ elements. And every other conjugacy class has exactly three elements in it. Since we need to account for $24$ other elements in the group, there are eight conjugacy classes with 3 elements each. In total, we get 11 conjugacy classes (the three with $1$ element each, the eight with $3$ elements each).

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So we use the class equation to find the number of conjugacy classes of $G$. Before I came back to this site, I discovered the paper on the five types of groups of order $p^3$, in which the first lemma just proved my first concern. As to the second, I thought I could avail of the regular representation to solve it, exactly contrary to the approach deploited by Issacs there in the book. And now I spot one other excellent, elementary, and somewhat opaque proof, on which I shall drill for more time. – awllower May 23 '12 at 15:04
Which proof do you find "somewhat opaque"? I'm not sure if we are using the class equation directly, just the fact that the conjugacy classes partition the group, and that the order of the conjugacy class of $x$ is equal to the index of its centralizer. – Arturo Magidin May 23 '12 at 16:20
Yes, you are absolutely right, and I start to perceive the principles behind this proof. Thanks. – awllower May 24 '12 at 5:32

Hint 1: if $G$ is non-cyclic group, then $G/Z(G)$ cannot be non-trivial cyclic

Hint 2: any group of order a prime is cyclic

Hint 3 (because of your remark): $G$ is abelian iff $Z(G)=G$

Yes, of course. In my answer, since

(1) $\,|Z(G)|=3\,$ and since

(2)$\,G/Z(G)\,$ is abelian,

then $\,G'\leq Z(G)\,$, and since it can't be $\,G'=1\,$ , then it must be $\,G'=Z(G)\,$ of order 3.

Now, it can't be $\,|Z(G)|=9, 27\,$ as then $\,G/Z(G)\,$ would be cyclic non-trivial, something that already was ruled out, or $\,G\,$ would be abelian, and the center cannot be trivial as $\,G\,$ is a p-group, as you mention...

It is not that the center of $\,G\,$ isn't trivial is "too elementary", but rather that it is a very important (or, as Amitzur would put it, the most important) characteristic of finite p-groups, so I suppose most of us assumed anyone asking what you did must already know this.

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About your hint 1: the conclusion also holds if $G$ is cyclic... in fact, for any group $G$, $G/Z(G)$ cannot be nontrivial cyclic. – Arturo Magidin May 23 '12 at 3:26
Do these hints include that the center of a p-group must not be trivial? I see not why through these hints? Per chance this is too elementary, well... – awllower May 23 '12 at 15:08
I address this comment in my second answer, which wasn't a comment as it came out too lengthy. – DonAntonio May 23 '12 at 16:07