Normal subgroups

1. Dec 28, 2008

math8

What is an example of a finite group G that has a subgroup H which is not normal but for which the index [G] is prime?

I was thinking about the alternating group A4 which has order 12, I know that the subgroups of A4 must have order 12, 6, 4, 3 , 2 or 1. But we need a prime index so I can eliminate subgroups of order 1,2 and 3 and 12. Also A4 does not have a subgroup of index 2, so I can also eliminate subgroups of order 6. So here I am looking for a subgroup of A4 of order 4.
What would that be? Or any other example that satisfies the requirements.

2. Dec 28, 2008

Citan Uzuki

Well, there is one subgroup of A_4 which has order 4, but unfortunately for your problem, it's also a normal subgroup of A_4. (Specifically, this subgroup is {(1), (12)(34), (13)(24), (14)(23)}). And this is the only subgroup of A_4 having order 4, so I'm afraid that using A_4 as the overgroup is a doomed approach.

Try letting G=S_5 and H be the obvious embedding of S_4 into S_5. I'm sure you can prove that the resulting subgroup isn't normal.

3. Dec 28, 2008

math8

Thanks, by embedding, do you mean that we just consider the permutations of the numbers {1,2,3,4} and assume that 5 is fixed? Also I was thinking maybe for proving that S_4 is not normal in S_5 to show that there exists an element g in S_5 such that gS_4g^-1 is not contained in S_4.
So I choose g=(12345), x=(1234) in S_4
hence gxg^-1=(1235) which is not in S_4. Is this valid?

4. Dec 28, 2008

Citan Uzuki

Yes, that's the embedding I was thinking of.

Your approach is right on, but your arithmetic is wrong. $gxg^{-1}(1) = gx(5) = g(5) = 1$, so $gxg^{-1}$ cannot possibly be (1235). (In fact, for your given choice, $gxg^{-1} = \text{(2345)}$).

5. Dec 28, 2008

math8

Oh yes, you're right, I guess I was doing g^-1xg instead of gxg^-1.

Thanks