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?
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).