Show that if $G$ is a finite group and $H_i$ are subgroups of $G$ with $[G:H_i]=2$ then $[G:\cap H_i]=$ some power of $2$.

My try:

Let the number of subgroups of $G$ be $H_1,H_2,\ldots ,H_m$

Its clear that each $H_i$ is a normal subgroup of $G$ and every $H_i$ has exactly two left/right cosets.

Let the left cosets of $H_1$ in $G$ be $H_1,g_1H_1$ ,that of $H_i$ in $G$ be $H_i,g_iH_i$ and so on.

Let $H=\cap H_i$

Now we know that $[G:H\cap K]\le [G:H][G:K]$ for any two subgroups $H,K$ of $G$.

Thus we have

$[G:H]\le 2^m$

Now I need to show only that $[G:H]\ge 2^m$

Now I understand that since $g_i\notin H_i\implies g_i\notin H$

hence we have at least $m$ cosets of $H$ in $G$ given by $g_1H,g_2H,\ldots g_mH$

But I need to find at least $2^m$

How can I do it?

Please give some hints

  • 1
    $\begingroup$ What you are trying to prove is stronger than asked, and in fact is not true. If you have $m$ subgroups of index $2$, there is no reason for the index of the intersection to be exactly $2^m$. You only need to show that it divides this number. Think about whether you can say something stronger than $[G:H\cap K]\le [G:H][G:K]$. (Think in terms of divisibility, rather than just magnitude.) $\endgroup$
    – verret
    Feb 14 '19 at 6:47
  • $\begingroup$ Think about the quotient group, as discussed below. $\endgroup$ Feb 14 '19 at 9:32

If $\vert G/H_i \vert =2$, then $\forall x \in G (x^2 \in H_i)$.

As you've noted, $\forall i H_i \lhd G$, so the quotient group $G/H_i$ has size two, and any representative $x$ of the quotient group's non-trivial element must satisfy $H_ix^2=H_i$, so $x^2 \in H_i$. (Of course, if $x \in H_i$, it's trivial that $x^2 \in H_i$.)

Since each $H_i \lhd G$, it follows that $\bigcap H_i \lhd G$.

Thus, every non-identity element of $G/ \bigcap H_i$ has order $2$.

If $p|~|G/ \bigcap H_i|$ for some odd prime $p$, then $G/ \bigcap H_i$ has a $p$-Sylow subgroup which has elements of odd order. We just finished proving that doesn't happen, so no odd prime divides $|G/ \bigcap H_i|$

Thus, $\vert G/ \bigcap H_i \vert = [G: \bigcap H_i] = 2^k$ for some $k$.

  • $\begingroup$ let $gH\in G/H$ how to show that $(gH)^2=H$ $\endgroup$
    – user596656
    Feb 14 '19 at 17:18
  • $\begingroup$ $gH\in G/H\implies g\notin H\implies \exists H_i $ such that $g\notin H_i\implies (gH_i)^2=H_i$ $\endgroup$
    – user596656
    Feb 14 '19 at 17:19
  • $\begingroup$ /Does that show that $gH$ has order $2$ how $\endgroup$
    – user596656
    Feb 14 '19 at 17:21
  • $\begingroup$ $gH$ has order $2$ (if $g \notin H$) because the quotient group $G/H$ is a group of order $2$, so its only non-identity element has order $2$. $\endgroup$ Feb 16 '19 at 10:02

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