Is H a normal subgroup...

[Genady]

... if it's a subgroup of a finite group G and contains exactly half the elements of G?

Edited December 28, 2024 by Genady

[KJW]

I can't speak about all groups G, but if there is a homomorphism that maps G to {1, –1}, then the kernel of the homomorphism (the subgroup of G that maps to 1) is a normal subgroup of G that contains exactly half the elements of G. In particular, the alternating group A_n of all even permutations of n objects, is a normal subgroup of the symmetric group S_n of all permutations of n objects.

It should be noted that Cayley's theorem states that every group is isomorphic to a subgroup of a symmetric group.

It occurred to me that the alternating group A_n where n is greater than or equal to 5 is a simple group of even order. That is, it has no normal subgroups other than the trivial group and the group itself. Thus, if the alternating group has any subgroups of order half that of the group, then the answer to your question is no. But there might not be any subgroups of order half that of the group.

 

Edited December 28, 2024 by KJW

[swansont]

Why is this in Brain Teasers? Seems like a math question.

[Genady]

2 hours ago, KJW said:

It occurred to me that the alternating group A_n where n is greater than or equal to 5 is a simple group of even order. That is, it has no normal subgroups other than the trivial group and the group itself. Thus, if the alternating group has any subgroups of order half that of the group, then the answer to your question is no. But there might not be any subgroups of order half that of the group.

Nice. So, you've proved that a simple group cannot have subgroups of order half that of the group. 😉

1 hour ago, swansont said:

Why is this in Brain Teasers? Seems like a math question.

Because I know the answer.

[swansont]

14 minutes ago, Genady said:

Because I know the answer.

Can it be figured out from the clues?

[Genady]

12 minutes ago, swansont said:

Can it be figured out from the clues?

I think so.

[KJW]

1 hour ago, Genady said:

Nice. So, you've proved that a simple group cannot have subgroups of order half that of the group. 😉

Hmmm. I did not realise THAT was what I had proven. From what I said above, proving that a map from G to {1, –1} is a homomorphism completes the proof that a simple group cannot have subgroups of order half that of the group.

 

[Genady]

7 hours ago, KJW said:

Hmmm. I did not realise THAT was what I had proven. From what I said above, proving that a map from G to {1, –1} is a homomorphism completes the proof that a simple group cannot have subgroups of order half that of the group.

 

You have shown that

10 hours ago, KJW said:

if the alternating group has any subgroups of order half that of the group, then the answer to your question is no.

Since the answer to the question is Yes, you have shown that the alternating group has no subgroups of order half that of the group.

[KJW]

10 hours ago, Genady said:

the answer to the question is Yes

At the time of my initial reply, I had not proved that the map from G to {1, –1} is a homomorphism. And instead of focusing on that proof, I digressed to a discussion about the alternating group. It was later that I devised a proof of the homomorphism based on Cayley tables. However, I subsequently discovered a direct proof that any subgroup of order half that of the group is normal in that group. The proof is that the left and right cosets of the subgroup must be equal, a defining property of normal subgroups.