What I've first done is show that $Aut(\mathbb Z / 24\mathbb Z)$ is isomorphic to $\mathbb Z_2\oplus\mathbb Z_2\oplus\mathbb Z_2$ due to it having 8 members and the greatest rank out of all of them is 2.

Now $M_2(\mathbb Z/3\mathbb Z)$ is isomophic to $\mathbb Z_3\oplus\mathbb Z_3\oplus\mathbb Z_3\oplus\mathbb Z_3$ (right?) so I need to find a homomorphism between those two groups.

I'm stuck here. Any help is appreciated.


You're looking for a non-trivial homomorphism $\phi$ of a group (let's call it $G$) of size 81 to one of size 8 (your isomorphisms are indeed correct!).

What can you say about $|Ker(\phi)|$, $|G/Ker(\phi)|$ and $|Im(\phi)|$ in terms of their factorisations?


Hint: If $\phi : M_2(\Bbb Z/(3))\to\operatorname{Aut}\Bbb Z/(24)$ is a homomorphism and $e\neq M\in M_2(\Bbb Z/(3))$, you must have $e = \phi(e) = \phi(M^{ord(M)}) = \phi(M)^{ord(M)}$. Hence, $ord(M)\mid \#\operatorname{Aut}\Bbb Z/(24) = 8$. But...

  • $\begingroup$ I am unable to understand to finish the path. Would you please complete it ? $\endgroup$ – Anjan3 Sep 13 '15 at 18:23
  • 1
    $\begingroup$ Ohh Ya ya I got it :-D $\endgroup$ – Anjan3 Sep 13 '15 at 18:23

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