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Let $G$ be a finite group of order $2^aq^b$ for odd prime $q$ and $a,b\geq 2$. Suppose $G$ has at least $2$ subgroups $A$ and $B$ of order $2^{a}$. I suspect that $A\cap B$ is a normal subgroup of size $2^{a-1}$. Any hint to the proof or give a counter example if you don't believe the proposition is true.

This question is answered with a counter example below.

INT
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  • Well, the edited question seems to be sounder, but: why do you suspect $;A\cap B;$ is (1) normal, and also (2) of size $;2^{a-1};$ ? This seems to be a lot to suspect if there's no more information. – DonAntonio Nov 14 '16 at 13:31
  • Anyway, that's what I want to apply it in my proof. But couldn't get a counter example why that doesn't hold, and couldn't come up with a proof either. – INT Nov 14 '16 at 13:32
  • I assume you're excluding the case $A = B$? – sTertooy Nov 14 '16 at 13:38
  • Yes, that's why I said at least two subgroups $A$ and $B$. – INT Nov 14 '16 at 13:38
  • @SteamyRoot Obviously, otherwise $;|A\cap B|=2^a;$ ... – DonAntonio Nov 14 '16 at 13:38

1 Answers1

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Counterexample:
Let $S_3$ be the symmetry group on $[1,3]$.
Let $G=S_3\times S_3$. $|G|=2^23^2$.
Let $\tau_1 = (1,2) \in S_3$ and $\tau_2 = (1,3) \in S_3$
Let $A$ be the subgroup generated by $(\tau_1,e)$ and $(e,\tau_1)$
Similarly let $B=\langle (\tau_2,e) , (e,\tau_2) \rangle$.
Then $A \cap B=\{e\}$.

This example can be extended to make $|A\cap B|=2^k$ for any $0 \leq k < a$.

Aydin Gerek
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