# 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 $2$

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?

• 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.) Feb 14 '19 at 6:47
• Think about the quotient group, as discussed below. 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 $$k$$.

• let $gH\in G/H$ how to show that $(gH)^2=H$
– user596656
Feb 14 '19 at 17:18
• $gH\in G/H\implies g\notin H\implies \exists H_i$ such that $g\notin H_i\implies (gH_i)^2=H_i$
– user596656
Feb 14 '19 at 17:19
• /Does that show that $gH$ has order $2$ how
– user596656
Feb 14 '19 at 17:21
• $gH$ 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 $2$. Feb 16 '19 at 10:02