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If $G$ group of order $52$ includes a normal group of order $4$ then $G$ is abelian.

I did like this $ |G|=52=2^2\cdot 13 $ let $H$ be normal group of order $4$. $n_{13}=1$ thus $G$ has a $K$ normal group order $13$. $|H|=4,|K|=13$, $H∩K=\{e\}$, $|H \cap K|=1$ , $H$ and $K$ finite groups $52=|HK|=|H|||K|/|H∩K|$ thus $HK=G$

I dont know either what to do next or if it's correct with this way

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    observation: the order of $K$ is prime; hence $K$ is cyclic (in particular, abelian).2012-10-28

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Since we have normal subgroups $H,K$ of $G$ with respective orders $4,13$, and since $H\cap K=\{e\}$ and $HK=G$, then $G\cong H\times K$. It suffices, then, to show that groups of order $4$ or $13$ must be abelian (the latter is very easy since a group of prime order is cyclic, and the former isn't too difficult), and that a product of abelian groups is abelian (again, not difficult).

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    I always like to say that small groups are abelian because formulating "is not abelian" requires a good handful of elements: If $ab\ne ba$, let $c=ab$, $d=ba$; then $1,a,b,c,d$ must be pairwise distinct (e.g. if $c=1$, then $b=a^{-1}$ and $ab=ba$).2012-10-28