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My question reads:

Let A, K be subgroups. Group G is called semidirect product of A and K if A $\trianglelefteq$ G, G=AK and A$\cap$K = < e >. Show that the groups are the semidirect product of two of its subgroups.

a) S$_3$

b) D$_4$

c) S$_4$

Now I am not sure if this is asking for a proof for each part or to directly pick two subgroups that are normal and then make sure the conditions for semidirect products are met. Also, doesn't this imply I need to show the subgroups I pick are normal? I need help picking these subgroups and from there I think it will be straightforward showing the other conditions are satisfied

For example for S3 could I pick the whole group itself?

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    I think the point of the exercise is to use $\textit{proper}$ subgroups. Otherwise, this is trivial. – 2017-02-21
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    @KenDuna well a hint to this problem is to consider normal subgroups and I think this makes more sense given that one of the conditions is that the subgroup be normal – 2017-02-21
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    My point is that every group is normal in itself. So $A = G$, $K = \{e\}$ would work for every group, $G$. This would make the question a silly one. So I think it is safe to assume that your teacher wanted $A$ and $K$ to be proper subgroups. With that said, a hint for $S_3$: $A = \langle (1 \ 2 \ 3) \rangle$ and $K = \langle (1 \ 2) \rangle$. – 2017-02-21
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    @KenDuna ah got it, yes those choices would be to easy. I was considering <(1 2)>. I guess my question is would I need to prove these subgroups I am picking are normal or can I just directly use them? – 2017-02-21
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    @KenDuna then K={ (1), (1 2)} but what would A look like – 2017-02-21
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    In (a), you want $A=\{(1),(1,2,3),(1,3,2)\}$. Try and do the others yourself. – 2017-02-21
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    @DerekHolt both subgroups do not need to be normal though, only one correct? – 2017-02-21
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    That's right, $A$ must be normal but not necessarily $K$. In fact if $K$ is normal then you get the direct product. – 2017-02-21
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    @DerekHolt okay, then I can simply take A and K and show that multiplying them I get G and there intersection is just < e> and I should be good? Then there is really no proof I am showing this is all just computation? I do not have to show these subgroups are normal? – 2017-02-22
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    But as I said before, the subgroup $K$ is not normal. For a complete proof you would need to prove that the subgroup $A$ is normal. – 2017-02-22
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    @KenDuna I am still confused by this problem, is there anyway you can show me how to do even just part a so I can do b and c? – 2017-02-22

2 Answers 2

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Per request: In $S_3$ let $A = \langle (1 \ 2 \ 3) \rangle$ and $K = \langle (1 \ 2) \rangle$. Note that the index of $A$ in $S_3$ is $2$ so we are guaranteed that $A$ is normal in $S_3$ (if you like, you can check normality with the definition).

Now our goal is to show that $S_3 = AK$. So let's just compute $AK$. There are $6$ quantities to compute, they are as follows:

\begin{align*} (1)(1) &= (1)\\ (1)(12) &= (1 \ 2) \\ (1 \ 2 \ 3)(1) &= (1 \ 2 \ 3)\\ (1 \ 2 \ 3)(1 \ 2) &= (1 \ 3)\\ (1 \ 3 \ 2)(1) &= (1 \ 3 \ 2)\\ (1 \ 3 \ 2)(1 \ 2) &= (2 \ 3) \end{align*}

You see every element of $S_3$ show up there, so $AK = S_3$.

Technically, this computation is unnecessary since $|AK| = \frac{|A||K|}{|A\cap K|}$. But I thought that it would be instructive in this case.

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    Would you mind if I did out part b here and you told me if it looks okay? I have just been struggling with this problem for days now and I want to make sure I understand it. – 2017-02-22
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In D$_4$, let A=< r>, K={e,s}. Checking that A is normal: by lagrange theorem, have index 2 thus A is normal.

Now showing D$_4$=AK We compute:

ee=e

es=s

re=r

rs=rs

r$^2$e=r$^2$

r$^2$s=r$^2$s

r$^3$e=r$^3$

r$^3$s=r$^3$s

Then, every element in D$_4$ shows up, so AK=D$_4$.

Then showing there intersection is < e > we can state it by looking at our sets.

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    Providing answers to your own questions is not only ok, but it is highly encouraged! With that said, $A \cap K \neq \{e\}$ in your example here ($r^2$ is in the intersection). Try using $K = \{e, s\}$. – 2017-02-22
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    Ok cool, I thought I wasn't allowed to do so. And yes! i just switched my K on my paper I will update it to see how it looks on paper and show you here. Thank You for the feedback. – 2017-02-22
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    Everything looks good. – 2017-02-22
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    Thank you! I am going to post c too I am just a bit unfamiliar with the elements of S4 as for my class we focus on S3. Would it be like S3 except we would have say (1 2 3 4) now? – 2017-02-22
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    Unfortunately that doesn't work. There are probably a number of ways to do the $S_4$ case, but I think the easiest way would be to set $A = A_4$ and $K = \langle (1 \ 2 ) \rangle$ where $A_4$ is the subgroup of all even permutations in $S_4$. In general $A_n$ is normal in $S_n$ (the index is 2). – 2017-02-22
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    Note that in $A_3 = \langle (1 \ 2 \ 3) \rangle \subseteq S_3$. So the approach is really the same as it was in $S_3$. – 2017-02-22
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    Hmm, so there is no simple way to list out the elements in S$_4$ like we did for S$_3$? What would A look like then set wise to get a sense of what elements to work with? – 2017-02-22
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    Writing out all the elements of $S_4$ or even $A_4$ is tedious because they have order $24$ and $12$ respectively. Every element of $S_n$ can be written as a product of transpositions. A transposition is an permutation of order 2 that switches two numbers and that is all it does. It can be verified that for any permutation, any time you express it as a product of transpositions, there is either always an even or always an odd number of them. A permutation is called even or odd based on this. $A_n$ is the set of even permutations. – 2017-02-22