Let $H_1,H_2$ be two distinct subgroups of finite group $G$ each of order $2$. Let $H$ be the smallest subgroup containing $H_1$ and $H_2$ . Then is it necessary that order of $H$ is amongst $2,4,8?$

I think that this is not true if we consider $S_3$ and $H_1=\{(1),(12)\}$ , $H_2=\{(1),(23)\}$ then smallest subgroup will be $S_3$ itself!

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    $\begingroup$ Yes, this is a valid counterexample to the statement. $\endgroup$ – Brian Dec 12 '13 at 4:22
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    $\begingroup$ As a matter of fact, given any positive integer $n$, there exists a group in which there are elements $x$ and $y$, each of order $2$, whose product $xy$ has order $n$. $\endgroup$ – Alexander Gruber Dec 12 '13 at 9:30
  • $\begingroup$ Alexander, great remark! One up from me! $\endgroup$ – Nicky Hekster Dec 12 '13 at 10:39

If $H_1$ or $H_2$ is normal, then your statement is true, in fact in that case $H=H_1H_2$ and $|H_1H_2|=4$. In general, the set $H_1H_2$ has 4 elements, since $|H_1H_2|=|H_1||H_2|/|H_1 \cap H_2|=2.2/1=4$, and of course $H_1H_2 \subseteq H$. So $H$ has at least 4 elements.


$H_1$={$1,a$}, $H_2=${$1,b$}. As the subgroups are distinct $a$ not equal to $b$. |$H_1\cup H_2|=3$. Note that $a,b$ are self inverse elements in $G$. Again if $H$ is the smallest group containing $H_1\cup H_2$ then $ab \in H$. Case 1. If $G$ abelian then $ab$ is self inverse element. so |H|=4. Case 2. If $G$ is not abelian $b^-1a^-1\in H$ . So |H|=5.


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