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I am stuck at another problem in my homework.

Find a non-Abelian group of size 48 such that the order of its elements are either 1, 2, 3 or 6.

I need some hints/tips to start on this problem as currently I don't know how I should approach this problem.

Also does anyone have any recommendations of any helpful websites that explains Abelian/non-Abelian groups/subgroups and how to find them? I think my understanding on this topic is really weak and would like to read up more.

Thanks a lot in advance for any help! :D

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$S_3\times\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2$.

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  • $\begingroup$ Nice and simple...+1 $\endgroup$ – DonAntonio Dec 5 '13 at 11:31
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    $\begingroup$ @DonAntonio: Thank you very much. $\endgroup$ – Boris Novikov Dec 5 '13 at 11:36
  • $\begingroup$ Very nice and simple. :) $\endgroup$ – mrs Dec 5 '13 at 13:06
  • $\begingroup$ @B.S.: Thank you very much. $\endgroup$ – Boris Novikov Dec 5 '13 at 13:13
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Start with the easiest non-abelian group, $S_3$, the permutation group of the set $\{1,2,3\}$. The elements of this group have order 1, 2 or 3. If you take the direct product of $S_3$ with a group of order $8$ where every element has order $2$, you're done.

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You are looking for an appropriate semi-direct product : If $|G| = 48$, then $G$ has a subgroup $H$ of order 16, and a subgroup $K \cong \mathbb{Z}_3$. So what you want is a non-trivial homomorphism $$ \varphi : K \to Aut(H) $$ which will give you a non-abelian semi-direct product.

Furthermore, the restriction on the orders of the elements of $G$ mean that $H$ cannot be cyclic, or be $\mathbb{Z}_4\times \mathbb{Z}_4$, for instance.

So perhaps you can try $H = \mathbb{Z}_2^4$. Then, $Aut(H) = GL_4(\mathbb{Z}_2)$. Now note that $3\mid GL_4(\mathbb{Z}_2)$, and hence there is an element in $Aut(H)$ of order 3. This will allow you to construct such a map $\varphi$.

Now it remains to be seen whether this semi-direct products satisfies the conditions you need - I think it should, but perhaps you can check that.

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