1
$\begingroup$

Let $G=\langle a,b\mid a^4=b^4=1, bab^{-1}=a^3\rangle$.

Please prove that $Aut(G)$ is generated by the automorphisms $$a\mapsto ab,\hspace{10pt} a\mapsto a^3,\hspace{10pt} a\mapsto ab^2, \hspace{10pt}b\mapsto a^2b \hspace{6pt}\text{and}\hspace{6pt}b\mapsto b^3.$$ and characterize the automorphisms group of $G$.

Explaintion for comment john mangual: Here we define the automorphisms on the the generatos $a$ and $b$. for example one of the automorphisms is $a\mapsto ab$, $b\mapsto b^3$. I think that the generators of $Aut(G)$ are the above form. Only i can not prove it.

Thank you

$\endgroup$
  • $\begingroup$ What are your thoughts on the problem? $\endgroup$ – Servaes Jun 20 '14 at 16:54
  • $\begingroup$ The subject "automorphisms" is very interesting for me and study this subject. $\endgroup$ – elham Jun 20 '14 at 16:59
  • $\begingroup$ :-) ${}{}{}{}{}$ $\endgroup$ – Mariano Suárez-Álvarez Jun 20 '14 at 18:11
  • 5
    $\begingroup$ @elham a group automorphism should show where both generators go, $a \mapsto ?, b \mapsto ?$ $$ . $$ how do you know $G$ is finite? $\endgroup$ – cactus314 Jun 20 '14 at 18:20
  • 1
    $\begingroup$ $G$ is a group of order 16 of nilpotency class two such that $|G'|=2$ and $|\dfrac{G}{Z(G)}|=4$ $\endgroup$ – elham Jun 20 '14 at 18:37
2
$\begingroup$

There is a natural isomorphism of $G$ with a semi-direct product $$\langle a \rangle \rtimes\langle b\rangle\simeq \mathbb{Z}/4\mathbb{Z}\rtimes \mathbb{Z}/4\mathbb{Z}$$ and where the action of b on $\langle a\rangle$ is given by $$\begin{array}{ccl} \langle a \rangle & \to & \langle a \rangle \\ a^i& \mapsto & a^{3i}=a^{-i}.\end{array}$$ Hence, $a^2$ commutes with $b$ and $a$ commutes with $b^2$. You can check that $$Z(G)=\langle a^2,b^2\rangle\simeq \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}.$$ (the inclusion is clear and neither $a$, nor $b$ nor $ab$ belongs to $Z(G)$). Moreover, the commutator subgroup of $G$, that you write $G'$, contains $a^{-1}bab^{-1}=a^2$ and is in fact equal to $$G'=\langle a^2\rangle \simeq \mathbb{Z}/2\mathbb{Z}.$$

Hence, you can check that $H=\{g\in G \mid g^2\in G'\}=\langle a,b^2\rangle$ (it is a maximal subgroup, and $b\notin H$).

Let us show that $\mathrm{Aut}(G)$ is generated by

$$\begin{array}{rcl} \tau_1\colon &G&\to& G\\ &a&\mapsto &ab^2\\ &b&\mapsto &b\end{array},\hspace{1cm} \begin{array}{rcl} \tau_2\colon &G&\to& G\\ &a&\mapsto &a^{-1}\\ &b&\mapsto &b\end{array},\hspace{1cm} \begin{array}{rcl} \tau_3\colon &G&\to& G\\ &a&\mapsto &a\\ &b&\mapsto &ab\end{array},\hspace{1cm} \begin{array}{rcl} \tau_4\colon &G&\to& G\\ &a&\mapsto &a\\ &b&\mapsto &b^{-1}\end{array}$$

Take $\tau\in \mathrm{Aut}(G)$. By definition of the subgroups, $\tau$ preserves $Z(G)$, $G'$ and $H$. Since $G'$ is of order $2$, $\tau(a^2)=a^2$.

Moreover, $\tau(a)$ is an element of $H$ whose square is $a^2$, so $\tau(a)\in \{a,a^{-1},ab^2,a^{-1}b^2\}$. Applying compositions of $\tau_1,\tau_2$ to any automorphism, we can assume that $a=\tau(a)$.

We then look at the action of $\tau$ on $G/Z(G)\simeq \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}.$ It fixes the class of $a$ and then either fixes the class of $b$ or exchanges it with the class of $ab$. Applying $\tau_3$ if needed, we can then assume that the class of $b$ is fixed, which means that $\tau(b)\in \{b,b^{-1},a^2b,a^2b^{-1}\}$. Hence, $\tau\in \langle \tau_3,\tau_4\rangle$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.