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Let $G$ is a finite group. $H\leq G$. $g^2 \in H$ for $\forall g (\in G)$ $\Rightarrow$ $H$ is normal subgroup of $G$

I tried to prove it by the different way using cosets. Let me show my proof.

Given the assumption I could get the cosets of $G$, $G = \{H, Hg\}$ (Here $g(\in G\setminus H)$ is the fixed element). Another $\forall x(\neq g)$ with $x \in G\setminus H$, I could also get $G=\{H,Hx\}$. Therefore My conclusion is $[G;H]=2$ since $Hg=Hx$ and $H$ is a subgroup of $G$, Finally $H$ is normal subgroup.

I can't find my errors in my proof. Is my proof right? If not please tell me which point I was wrong. Best regards.

Shaun
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kechang lee
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  • Yes, you are right, since $g^2 \in H$ then $gH$ must be the same as $Hg$. – Salcio Nov 15 '22 at 02:38
  • Use the following: Any subgroup containing the commutator subgroup is normal, a normal subgroup contains the commutator subgroup if and only if the quotient group is abelian, the subgroup generated by the squares is normal, and any group all of whose elements square to the identity is abelian. – Geoffrey Trang Nov 15 '22 at 02:38

3 Answers3

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How did you get that there are only two cosets? It is certainly not true when $H$ is the trivial subgroup in any group where every element has order two, like $C_{2}\times C_{2}$. Instead, take any $g \in G, h \in H$ then $ghgh=(gh)^2 \in H$. Multiplying to the right by $h^{-1}$ and by $g^{-2}$, both of which are in $H$, we get $ghg^{-1}=(ghgh)h^{-1}g^{-2} \in H$

Iker
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Maybe this is simpler - take $g \in G$. Then $g^2 = h$, for some $h \in H$. In words, $g=hg^{-1}$. Multiply both sides by $g$ to get $g^2= ghg^{-1}$. But this is $h$ so it is in $H$.

Salcio
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    You are not proving that $ghg^{-1}\in H$ for all $g\in G$ and $h\in H$. You only proved that given $g$, there is a particular $h\in H$ such that $ghg^{-1}\in H$, and that is not enough. – Arturo Magidin Nov 15 '22 at 04:38
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Look at left and right cosets: you get by same argument as yours with $g \notin H$, $[G:H] = 2$: $H \cup gH = G$ and $H \cup Hg = G$ since cosets are disjoint, we get $gH = Hg$ =>$gHg^{-1} = H$ => $H$ is normal subgroup.

Balaji sb
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