# Thread: S4 subgroup of order 12

1. ## S4 subgroup of order 12

Show that S4 (the symmetric group of degree 4) has a unique subgroup of order 12.

I know that A4 is that subgroup but I'm not really sure how to show that it is the unique subgroup. Help?

2. ## Re: S4 subgroup of order 12

Hi,
I'm tired of trying to write latex in this editor. So the hints are in an attachment. If you have problems, post them.

thank you!

4. ## Re: S4 subgroup of order 12

We can approach this another way:

Consider the homomorphism $\displaystyle \text{sgn}:S_4 \to \{-1,1\}$.

If $\displaystyle H$ is ANY subgroup of $\displaystyle S_4$, then $\displaystyle H \cap A_4$ is also a subgroup of $\displaystyle S_4$. Restricting the homomorphism sgn to this subgroup yields a homomorphism from $\displaystyle H \cap A_4$ to {-1,1}.

Since there are only two possibilities for the image of this homomorphism, we have either:

$\displaystyle \text{sgn}(H \cap A_4) = \{1\}$, which implies that $\displaystyle H$ is a subgroup of $\displaystyle A_4$ (why?), or:

$\displaystyle \text{sgn}(H \cap A_4) = \{-1,1\}$, which implies that $\displaystyle H \cap A_4$ is of index 2 in $\displaystyle H$.

Now if $\displaystyle |H| = 12$, the first possibility leads to $\displaystyle H = A_4$. So if $\displaystyle H \neq A_4$, we must have that $\displaystyle |H \cap A_4| = |H|/[H: H \cap A_4] = 12/2 = 6$.

Now $\displaystyle H \cap A_4$ is thus a subgroup of $\displaystyle A_4$ of order 6. There are two possibilities:

a) This subgroup is cyclic, but $\displaystyle A_4$ has no elements of order 6, which leaves us with:

b) $\displaystyle H \cap A_4 \cong S_3$.

However, if $\displaystyle A_4$ has a subgroup isomorphic to $\displaystyle S_3$, this subgroup would contain 3 elements of order 2. However, the (only) 3 elements of order 2 in $\displaystyle A_4$, namely:

(1 2)(3 4), (1 3)(2 4), (1 4)(2 3) generate a subgroup of order 4, and $\displaystyle S_3$ has no such subgroup (it cannot, for 4 does not divide 6).