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We can prove that both of:

$S_3=\mathbb Z_3\rtimes\mathbb Z_2$ and $\mathbb Z_6=\mathbb Z_3\rtimes\mathbb Z_2$

So two different groups (and not isomorphic in examples above) can be described as semi-direct products of a pair of groups ($\mathbb Z_3$ and $\mathbb Z_2$). I hope my question does make sense:

Is there any group which can be described as semi-direct products of two different pair of groups?

Thanks.

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What have you tried? (I assume this isn't homework, but I still think questions like this make valuable self-imposed exercises and it is not clear to me that asking them on math.SE is the most productive thing to do before doing some work on them.) –  Qiaochu Yuan Sep 16 '12 at 19:25
    
@QiaochuYuan: Thanks for what you noted me. I got more from them. Actually, I didn't know how to ask this kind of question here. I'll try to do what you said. Thanks again for your time. –  B. S. Sep 17 '12 at 11:24
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4 Answers

up vote 6 down vote accepted

Here is a non-trivial example the comes up in practice:

Whenever $p$ is a prime number, there are two non-abelian groups of order $p^3$:

  • the semidirect product of $\mathbb{Z}/p\mathbb{Z}$ acting as $\big(\begin{smallmatrix} 1 & 1 \\. & 1 \end{smallmatrix}\big)$ on $\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z}$, and
  • the semidirect product of $\mathbb{Z}/p\mathbb{Z}$ acting as $x\mapsto x^{1+p}$ on $\mathbb{Z}/p^2\mathbb{Z}$.

However, when $p=2$ the unthinkable happens: the two groups are both isomorphic to the dihedral group of order 8!

The “dihedral” aspect of the group is mostly the second semi-direct product, with the reflections acting on the group of rotations. However the first is very important as the Sylow 2-subgroup of the simple group GL(3,2) of order 168.

When $p$ is an odd prime, the two groups are very different. The first has exponent $p$ (all elements have order $p$) while the second visibly has an element of order $p^2$.

This sort of behavior is not too uncommon in p-groups of larger order and is one difficulty in “naming” them, even the ones who look like they should have names.

(The other examples given apply equally well as “counterexamples” to the fundamental theorem of arithmetic: $16 = 2 \times 8 = 4 \times 4$, $12 = 4 \times 3 = 6 \times 2$, or even $4=1\times 4 = 2 \times 2$. The example given here is not of this form, as $\mathbb{Z}/p^2\mathbb{Z}$ is not a non-trivial semi-direct product.)

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Such a group of smallest order is $D_8$, the Dihedral group of order 8.

Write $D_8=\langle x,y\colon x^4=y^2=1, y^{-1}xy=x^{-1}\rangle=\{1,x,x^2,x^3, y,xy,x^2y,x^3y \}$.

  1. $H=\langle x\rangle$,$K=\langle y\rangle$, then $D_8=H\rtimes K\cong C_4\rtimes C_2$.

  2. $H=\langle x^2,y\rangle$, $K=\langle xy\rangle$, then $D_8=H\rtimes K\cong (C_2\times C_2)\rtimes C_2$.

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Only if you don't allow direct products to be semidirect products! –  Qiaochu Yuan Sep 17 '12 at 16:30
    
of course! for abelian groups, its easy to find examples; $C_3\timec C_2\cong C_6\times C_1$!!! –  Marshal Kurosh Sep 18 '12 at 3:16
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For one example, we have $\mathbb{Z}_6 \times \mathbb{Z}_2 \cong \mathbb{Z}_3 \times \mathbb{Z}_{2^2}$, since direct products are a case of semi direct products with trivial action.

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Sure. Take $G = C_2^4$, which is the direct product of $C_2$ and $C_2^3$ but also the direct product of $C_2^2$ and $C_2^2$. An even less interesting example is $G = C_2^2$, which is the direct product of $1$ and $C_2^2$ but also the direct product of $C_2$ and $C_2$. I assume you wanted more "interesting" examples, though.

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