Prove/Disprove that $A_4\cong D_3×\mathbb Z_2$:

I can't seem to find a way to disprove it (Something in my mind it telling me it can't be true). But the order of both is 12, both are non cyclic, non abelian.

Can I simply say that the order of an element in $D_3$ is 3 and combined with $\mathbb Z_2$ it will have order 6, when in $A_4$ the maximum order of an element is 3?

2)Another Question:

Consider the set $H$:={$e,(24)(13),(14)(23),(34)(12)$ }

Is $H\cong \mathbb Z_4$? Prove it's normal in $S_4$

I don't know how to show/disprove isomorphism (I think it's false). It is normal because it's a union of conjugacy classes and the $geg^{-1}=e\in S_4$

  • 5
    $\begingroup$ Yes, $A_4$ has no element of order $6$, while $D_3\times Z_2$ does. You can use the same idea with the second question: what are the orders of the elements of $H$ and of $Z_4$? $\endgroup$ – James Apr 20 '15 at 15:19
  • $\begingroup$ Ok yeah it does work. H has an element of order 2, and fir Z4 its 4. I take it the proof of normality is correct? $\endgroup$ – GRS Apr 20 '15 at 15:30
  • $\begingroup$ I think that the best way to argue this is to just note that $D_3 \times Z_2$ has a normal subgroup isomorphic to $D_3$ (direct products always contain normal subgroups isomorphic to their factors), and $A_4$ does not. $\endgroup$ – stochasm Apr 20 '15 at 15:38
  • $\begingroup$ Nick: it's not just that $H$ has an element of order 2; rather, all elements of $H$ are of order 2. (Why?) $\endgroup$ – Steven Stadnicki Apr 21 '15 at 23:51
  • $\begingroup$ Is it because it has to be a union of the same conjugacy class (the same cycle type) to be normal? $\endgroup$ – GRS Apr 25 '15 at 15:20

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