# How to show that Klein four-group is a normal subgroup of the alternating group $A_4$

I want to show that the Klein four-group is a normal subgroup of the alternating group $A_4$.

I am using the information in this link, that shows explicitly $A_4$, and Klein four-group as a subgroup.

I know that there is the direct way, by definition, but is there a way that does not require actually multiplying so many permutations ?

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@Chris Eagle - forgot it, I added it to the post –  Belgi Mar 17 '12 at 21:50

The elements of the Klein $4$-group sitting inside $A_4$ are precisely the identity, and all elements of $A_4$ of the form $(ij)(k\ell)$ (the product of two disjoint transpositions).

Since conjugation in $S_n$ (and therefore in $A_n$) does not change the cycle structure, it follows that this subgroup is a union of conjugacy classes, and therefore is normal.

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I will accept this after the timer allows me. I forgot that conjugation does not change the cycle structure.. Thank you! –  Belgi Mar 17 '12 at 21:55
@Belgi: Also, remember that in general it suffices to show that $ghg^{-1}\in H$ and $g^{-1}hg\in H$ for every $h$ in a generating set for $H$ and every $g$ in a generating set of $G$ in order to show that $H$ is normal in $G$; so you often don't have to compute all possible $ghg^{-1}$, just some. –  Arturo Magidin Mar 17 '12 at 22:25
thank you again. Is there a simple way to see why does this subgroup is closed under multiplication ? –  Belgi Mar 17 '12 at 22:44
@Belgi: Which group? The Klein $4$-group inside $A_4$? Noting that $(12)(34)\circ(13)(24) = (14)(23)$ is sufficient, together with the fact that each of them is its own inverse. –  Arturo Magidin Mar 17 '12 at 22:46
Pretty solution! –  Bombyx mori Jul 1 '13 at 3:28

You do not have to compute the entire Cayley table of a group to get an isomorphism. I think that this is what you are struggling with so I'll put a little of detail.

First of all, there are only two groups of order $4$ : the cyclic group of order $4$ and the Klein group, it is quite easy to see it. Let $G$ be a group of order $4$ : by Cauchy's theorem, since $2$ divides $4$ and $2$ is prime, there exists an element of order $2$ in $G$. Let $a$ be this element and $H = \langle a \rangle$. Since $[G:H] = 2$, $H \triangleleft G$, hence for all $g \in G$, $gag^{-1} \in H$, but since $a \neq 1$, we must have $gag^{-1} = a$, that is, $a$ commutes with every element of $G$.

Consider an element outside $H$, call it $b$. If it has order $4$, then $G = \langle b \rangle \cong C_4$, the cyclic group of order $4$. If not, then since the order of an element divides the order of the group, and $b \neq 1$, then $b$ must have order $2$. But then the fourth element cannot have order neither $1$ or $4$, hence it must have order $2$ too. Hence every non-trivial element of $G$ has order two, and using the argument above, they commute with each other ; this gives you the group $C_2 \times C_2$, which is precisely the Klein group, because we must have $ab = c$, $ac = b$ and $bc = a$ (for obvious reasons, because other possibilities lead to contradictions).

Now the subgroup of $A_4$, namely $K=\{(1), (12)(34),(13)(24),(14)(23) \}$ is a subgroup of order $4$. Since it is not cyclic, it is isomorphic to the Klein group. Conjugation in $S_n$ does not change cycle structure, so that in particular it does not do that in $A_n$. This means that this subgroup is normal, because $gKg^{-1} \subseteq K$, which is an equivalent condition for normality of a subgroup.

Hope that helps,

P.S. : Maybe I used sledgehammers to classify groups of order $4$, but I just threw out the first ideas that came to mind ; if you want to simplify them by commenting feel free.

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Why would you want to classify groups of order 4? I don't think it is relevant. Furthermore, the end of the argument is just a rewrite of Arturo's answer... –  M Turgeon Mar 17 '12 at 22:06
Because then looking at this subgroup $K$ of $A_4$ you know it is isomorphic to $C_2 \times C_2$ by noticing that there is no element of order $4$ in this subgroup ; instead of computing the isomorphism by hand (which OP asked for, i.e. "is there a way without multiplying so many permutations"), I only use the fact that I know which are the groups of order $4$. The end of the argument is just a re-write indeed, but it is not the point of my answer. –  Patrick Da Silva Mar 17 '12 at 22:18
The question was not "what is the isomorphism type of this subgroup", but rather "how can I check it is normal without computing 'so many' products?" (i.e., without having to check $ghg^{-1}\in H$ for every $h\in H$ and possibly every $g\in G$). –  Arturo Magidin Mar 17 '12 at 22:24
I didn't understand the question well enough without the link, because I didn't know that OP knew that the Klein group was actually isomorphic to the subgroup, which was the question I answered. Now that OP added his link I took a look and obviously I answered a different question than his. –  Patrick Da Silva Mar 17 '12 at 22:29