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I have made some headway with the proof, but I can't quite finish it off. Please could I have some help? Please note that at no point are Sylow Theorems to be used during this proof.

Let $G$ be a group such that $|G| =28$. We are given $H$, such that $|H|=4$, and $H$ is a normal subgroup of $G$. Previously, I have proven that $G$ must also contain a normal subgroup, $K$, where $|K|=7$. This was done without Sylow. I noticed that $H$ must be isomorphic to $C_4$ or $C_2\times C_2$ because these are the only groups of order 4, up to isomorphism. Since I'm trying to show $G$ is abelian, I guessed that $G$ will be $C_{28}$ or $C_2 \times C_{14}$.

To try and show this, I started using the Direct Product Theorem. In either case of the identity of $H$, $H \cap K = e$, because $H$ will not contain any elements of order 7, and all the elements of $K$ are order 7 apart from the identity. Also, $H$ and $K$ are normal, so their elements commute with each other: For $h \in H$ and $k \in K$, $(khk^{-1})h^{-1} \in H$ and $k(hk^{-1}h^{-1}) \in K$ means $khk^{-1}h^{-1} = e$.

But I cannot work out how to show any element of $G$ is the product of elements in $H$ and $K$. Should I perhaps consider the order of elements in $G$? When $H$ is $C_4$, $G$ will be $C_{28}$ and so must contain an element of order 28.

Any help is very much appreciated.

citadel
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John Smith
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    Since $H$ and $K$ are normal, the (element-wise) product $HK$ is a subgroup of $G$ containing both $H$ and $K$ (it would have sufficed that one of the two is normal). Now, what is the smallest possible order of a group containing $H$ and $K$? – Sebastian Schoennenbeck Nov 21 '17 at 15:15
  • If you have proved that $H$ and $K$ commute elementwise, what is left to prove? Note that both $H$ and $K$ are abelian. – lhf Nov 21 '17 at 15:18
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    More directly, $c\colon g \mapsto c_g$, where $c_g(h) = ghg^{-1}$ is a homomorphism $G \to \operatorname{Aut} H$. We know $H \subset \ker c$, and $7 \nmid [\operatorname{Aut} H :1]$. – Daniel Fischer Nov 21 '17 at 15:19
  • Ahh, I follow. That answers my question, so thank you very much, @SebastianSchoennenbeck. But why is it that one of $H$ and $K$ being normal means their product group is a subgroup of $G$? – John Smith Nov 21 '17 at 15:20
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    In natural notation (let's say $H$ is the normal subgroup): $hkh'k' = hkh'(k^{-1}k) k'=(h kh'k^{-1}) (kk')$, the first factor being in $H$ since it is normal and the latter factor being in $K$. The identity element is clearly contained in $HK$ and inverses go pretty much analogously to the product. – Sebastian Schoennenbeck Nov 21 '17 at 15:26
  • @DanielFischer Just to clarify, $[\operatorname{Aut} H :1]$ means $\circ (\operatorname{Aut} H)$. – Vinay Deshpande Jul 02 '20 at 04:33
  • @VinayDeshpande Yes, it's the order of $\operatorname{Aut} H$. Generally, $[A : B]$ is a notation for the index of the subgroup $B$ in the group $A$, and the trivial subgroup is often denoted by $1$, thus $[A : 1]$ is the index of the trivial subgroup in $A$, which equals the order of $A$. – Daniel Fischer Jul 20 '20 at 13:10

3 Answers3

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By Cauchy there is an element of order $7$, and hence a subgroup $K\cong C_7$. Since $H$ is normal, $HK$ is a subgroup of $G$ order $|HK|=\frac{|H||K|}{|H\cap K|}=\frac{4\cdot 7}{1}=28$, so $HK=G$. Therefore, $G\cong K\ltimes H$. But $K$ acts trivially on $H$, regardless of the structure of $H$ and even less of that of $\operatorname{Aut}(H)$ (not even its order, see e.g. here, in particular the last paragraph). So, $G$ is isomorphic to the direct product of abelian groups, and hence it is abelian in turn.

citadel
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$G$ is the semidirect product of $H$ and $K$, now, up to isomorphisms there are two groups of order 4,$Z/4Z$ and $Z/2Z\times Z/2Z$,the group of Automorphisms of $Z/4Z$ is isomorphic to $Z/2Z$,and the automorphism group of $Z/2Z\times Z/2Z$ is isomorphic to $S_{3}$, in both cases the only homomorphism from $K=Z/7Z$ to $Aut(H) $ is the trivial homomorphism, So in fact $G$ is the direct product of $H$ and $K$, since both are abelian, we deduce that $G$ is Abelian. Of course semidirect product can be avoided, since u proved that both subgroups are normal, indeed $G=HK$ by the formula for product, and the intersection is trivial. Let $x\in H, y\in K$, then $xyx^{-1}y^{-1}\in H \cap K$ because $xyx^{-1} \in K$ since K is normal, so $xyx^{-1}y^{-1}\in K$ the same reasoning shows that it belongs also to $H$, since the intersection is trivial $xyx^{-1}y^{-1}=1$,i.e $xy=yx$, this shows that $G$ is isomorphic to the direct product of $H$ and $K$, hence it is Abelian.

belkacem abderrahmane
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If $o$ denotes order, $o(HK)=(o(H)o(K))/o(H\cap K) $.

Here, $o(HK)=7*4=28$ ,so every element $g$ of the group $G$ is in $HK$. So it can be written as $g=hk$.

Integrand
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hasan kibria
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