MHB If G is the union of 3 of its subgroups, show that they each have index 2

[oblixps]

Let G be a finite group and let G = H_1 \cup H_2 \cup H_3. Show that [G: H_i] = 2 for i = 1, 2, 3.

There was a hint for this question saying to first prove that at least one of the subgroups has index 2 in G. So far I am not sure how to even start this problem. I know that the orders of H_1, H_2, H_3 must divide the order of G, but this doesn't give much information. I was thinking of trying to show that the order of G must have a factor of 2, since |G| = [G: H_i]|H_i| but with only the information given, I have no idea how to go about it.

Could someone offer a hint or two on how to proceed? thanks.

 

[Amer]

I think there is something missing from the question, as you said [G] should also divide G order what about a group with order dose not divide 2, odd order for example
\mathbb{Z}_{9} with addition has odd order.
G order should be even

 

[Opalg]

Let G be a finite group and let G = H_1 \cup H_2 \cup H_3. Show that [G: H_i] = 2 for i = 1, 2, 3.

There was a hint for this question saying to first prove that at least one of the subgroups has index 2 in G.

To prove the hint, suppose (in order to get a contradiction) that each of the subgroups has index greater than 2. Then $|H_i| \leqslant \tfrac13|G|$ (for $i = 1, 2, 3$). In other words, each of the subgroups contains at most 1/3 of the elements of $G$. But the subgroups have nonempty intersection, because the identity element is in each of them. It follows that their union does not contain enough elements to span the whole of $G$.

 

[oblixps]

thanks for the reply. I now pick the subgroup of index 2 to be H_1 so since |G: H_1| = 2, i know that |H_1| = |G|/2 and I was able to deduce that G = H_{1}H_{i} and |H_i : H_{1} \cap H_{i}| = 2 for i = 2, 3 (this was another hint given with the problem as well). I'm having a hard time thinking of what to do next.

I have tried writing |H_2| = 2|H_{1} \cap H_{2}| and |H_3| = 2|H_{1} \cap H_{3}| and i tried using the inclusion exclusion principle so that |G| = |H_1| + |H_2| + |H_3| - |H_{1} \cap H_{2}| - |H_{1} \cap H_{3}| - |H_{2} \cap H_{3}| + |H_{1} \cap H_{2} \cap H_{3}|. Substituting everything i know so far, i have |G|/2 = |H_{1} \cap H_{2}| + |H_{1} \cap H_{3}| - |H_{2} \cap H_{3}| + H_{1} \cap H_{2} \cap H_{3}|. but the trouble is i don't know anything about the last 2 terms. is this the way to go? or is there a better way of using this 2nd hint?

 

[Opalg]

thanks for the reply. I now pick the subgroup of index 2 to be H_1 so since |G: H_1| = 2, i know that |H_1| = |G|/2 and I was able to deduce that G = H_{1}H_{i} and |H_i : H_{1} \cap H_{i}| = 2 for i = 2, 3 (this was another hint given with the problem as well). I'm having a hard time thinking of what to do next.

I have tried writing |H_2| = 2|H_{1} \cap H_{2}| and |H_3| = 2|H_{1} \cap H_{3}| and i tried using the inclusion exclusion principle so that |G| = |H_1| + |H_2| + |H_3| - |H_{1} \cap H_{2}| - |H_{1} \cap H_{3}| - |H_{2} \cap H_{3}| + |H_{1} \cap H_{2} \cap H_{3}|. Substituting everything i know so far, i have |G|/2 = |H_{1} \cap H_{2}| + |H_{1} \cap H_{3}| - |H_{2} \cap H_{3}| + H_{1} \cap H_{2} \cap H_{3}|. but the trouble is i don't know anything about the last 2 terms. is this the way to go? or is there a better way of using this 2nd hint?

You have identified the most important fact, which is that $|H_i : H_{1} \cap H_{i}| = 2$. If I can express it informally, that means that half of the elements of $H_i$ are in $H_1$ ( for $i=2,3$). The other half are in the coset $G\setminus H_1$, which is half the size of $G$. If either $H_2$ or $H_3$ has index $\geqslant 3$ then that $H_i$ is at most 1/3 the size of $G$, so the half of it in the coset $G\setminus H_1$ is at most 1/6 the size of $G$. As before, that will not be enough for the two groups $H_2$ and $H_3$ together to cover the whole coset.