# Automorphisms of a finite non-abelian group

Sia $G$ un gruppo non abeliano di ordine $21$, quanti automorfismi possiede un gruppo siffatto?

Translation (by Davide): Let $G$ be a non-abelian group of order $21$. How many automorphisms does such a group possess?

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Sarebbe meglio tradotto in inglese, così non solo gli italiani ti possano rispondere. Grazie. –  Davide Giraudo Dec 22 '11 at 10:07
@Davide: it's actually alright, if the asker's English isn't that good. But since you seem to be able to understand what he wants, could you maybe edit the question to include an English translation? –  Guess who it is. Dec 22 '11 at 10:10
...on that note, "how many automorphisms does a nonabelian group of order 21 have?" is essentially the query, no? –  Guess who it is. Dec 22 '11 at 10:12
"in this manner"... –  David Mitra Dec 22 '11 at 10:16
Lots of nice answers. :) I hope somebody can do an Italian translation for the OP. –  Guess who it is. Dec 22 '11 at 16:17

A different solution.

Notice first that $G$ has trivial center, so $|\operatorname{Inn}(G)| = |G/Z(G)| = 21$. Thus $|\operatorname{Aut}(G)|$ is a multiple of $21$. Then bound $|\operatorname{Aut}(G)|$ above by $42$, so $|\operatorname{Aut}(G)| = 21$ or $|\operatorname{Aut}(G)| = 42$. Finally, you can find an automorphism that is not inner, and thus it turns out that the answer is $42$ (as usual).

Here are details for the last part. By Cauchy's theorem we find non-identity elements $x$ and $y$ in $G$ such that $x^3 = y^7 = 1$. Since $\langle y \rangle \trianglelefteq G$, we have that $x^{-1}yx = y^{\alpha}$ where $1 \leq \alpha \leq 6$. Then $y = x^{-3}yx^3 = y^{\alpha^3}$, and so $\alpha^3 \equiv 1 \mod 7$. Therefore $\alpha = 2$ or $\alpha = 4$, because $G$ would be cyclic in the case of $\alpha = 1$. Let's assume $\alpha = 2$ for clarity, the other case is similar (and the group would still be the same up to isomorphism). Thus $x^{-1}yx = y^2$.

It is not difficult to show that the elements of $G$ are of the form $x^ky^l$, where $0 \leq k \leq 2$ and $0 \leq l \leq 6$.

Let $\phi: G \rightarrow G$ be an automorphism of $G$. Because $G$ is generated by $x$ and $y$, we have that $\phi$ is completely determined by the values of $x$ and $y$. Now $\phi(y)$ and $y$ have the same order, so $\phi(y) = y^j$, where $1 \leq j \leq 6$. Similarly $\phi(x)$ and $x$ have the same order, so $\phi(x) = xy^i$ or $\phi(x) = x^2y^i$, where $0 \leq i \leq 6$.

Could it be that $\phi(x) = x^2y^i$? Then $\phi(xy)\phi(xy) = x^2y^{5i+5j}$ and $\phi(xyxy) = \phi(x^2y^3) = xy^{5i+3j}$. Therefore $y^{2j} = 1$, but this means that $y^j = 1$, which is not possible.

Thus there are $7$ possibilities for the value of $\phi(x)$. Since there are $6$ possibilities for the value of $\phi(y)$, we have that $|Aut(G)| \leq 42$. Finally we need to find an outer automorphism. Because $y$ and $y^3$ are not conjugate, we could try $\phi(x) = x$ and $\phi(y) = y^3$.

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Such a group is the semidirect product $C_7\rtimes C_3$. Let $C_7=\langle x\rangle$ and $C_3=\langle y\rangle$. Here $y$ acts by $x^y=x^2$. Then an automorphism $\alpha$ must send $x$ to some power of $x$ (because $\langle x\rangle$ contains all six elements of order $7$); there are $6$ choices there.

$\alpha$ must then send $y$ to some $x^iy^j$ Such that $x^{y^j}=x^2$; since we can consider $j\pmod{3}$, we need $j=1$. There are $7$ choices for $i$ (from $0$ to $6$), and so there are at most $42$ automorphisms of our group.

But every such choice yields an automorphism. There are many ways to show this; I like taking a presentation for your group to be $\langle x,y\ |\ x^7=y^3=1, x^y=x^2\rangle$, and then simply checking that for each of the $42$ choices for $\alpha$, $\alpha(x)$ and $\alpha(y)$ generate your group, and $\alpha(x)^7=\alpha(y)^3=1$, with $\alpha(x)^{\alpha(y)}=\alpha(x)^2$. Finiteness then guarantees that all such choices give automorphisms.

It's also not hard to show the automorphism group is the holomorph of $C_7$, namely $C_7\rtimes C_6$. Here the generator for the $C_7$ factor is the map sending $x$ to $x$ and $y$ to $xy$. The generator for the $C_6$ factor is the map sending $x$ to $x^3$ and $y$ to $y$.

All of this works in more generality as well. For the finite centerless group $C_m\rtimes C_n$, there are $m\phi(m)$ automorphisms, and the automorphism group is the holomorph $C_m\rtimes C_{\phi(m)}$.

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Steve why $\alpha$ must send y to $x^iy^j$ such that $x^{y^j}=x^2$ ? –  WLOG Dec 22 '11 at 14:46
@stef: $\alpha$ needs to preserve the relations in the group - it's supposed to be an isomorphism after all! I talk more about this in the third paragraph. –  user641 Dec 22 '11 at 18:06

There may be easier ways to do this, but here is one. Let $A$ be the automorphism group of $G$. Note that $G$ has seven Sylow $3$-subgroups and $14$ elements of order $3$. Any two different Sylow $3$-subgroups generate $G,$ so no non-identity element of $A$ can fix more than two elements of order $3$ in $G.$ On the other hand, any element of $A$ which fixes an element of order $3$ has to fix the inverse of that element. Hence every non-identity element of $A$ fixes $0$ or $2$ elements of order $3$ in $G.$ Let $\theta$ be the permutation character of $A$ afforded by its action on the elements of order $3$ in $G$. Then $\theta(\theta- 2 \times 1_A)$ vanishes on all non-identity elements of $A$ (where $1_A$ denotes the trivial character). It is a generalized character of degree $168$, so that $|A|$ divides $168$ (taking the inner product with $1_A$).

\medskip On the other hand, let $F$ be a Sylow $7$-subgroup of $A$. Let $a$ be an automorphism of order $2$ in $A$. Suppose that $a$ centralizes $F$ (note that $G$ has trivial centre, so is isomorphic to ites inerr automorphism group, so may be regarded as a normal subgroup of $A$). Notice that $C_G(F) = F.$ Now we have $[\langle a \rangle,F,G] = [F,G,\langle a \rangle ] =1,$ so that by the three subgroups lemma, $[G,\langle a \rangle, F] = 1.$ But $C_G(F) = F,$ so that $[G,\langle a \rangle] \leq F$ and $[G,\langle a \rangle, \langle a \rangle] = 1.$ Since $|G|$ is odd, standard results on coprime automorphisms show that $[G,\langle a \rangle ] = 1,$ a contradiction since $a$ has order $2$.

\medskip Hence a Sylow $2$-subgroup of $A$ acts faithfully as a group of automorphisms of $F,$ so is cyclic of order (at most) $2$. Hence $|A|$ divides $42$. But $G$ is isomorphic to a subgroup of the holmorph of $F,$ which has order $42$, so $|A| = 42.$

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