# Symmetric subgroups

Let $A =$ the symmetric group in $S_4$. Consider the following subgroup of $A$: $$J=\langle(1234),(12)(34)\rangle$$ Find $|J|$.

I believe that the subgroup must contain the identity and that $(1234)$ is a cyclic subgroup of order $4$ so $4$ must divide the order of $A$ (Cayley's and Lagrange). That gives us $3$ elements and that $|J|$ must be $4$, $8$, $12$, or $24$.

I then multiplied $(1234)*(12)(34)$ to get $(13)$ and that's another element. Also, I know the inverse of $(1234)$ must be in the subgroup as well. I've narrowed the possibilities down to $|J|$ just be $8$, $12$, or $24$. My guess is the answer is $8$ but I'm not completely sure of my calculations and I don't know how to find more elements to prove this.

• It may be easier to think of this geometrically, consider the map $J \to D_4$ where we map generators to generators. ($D_4$ is the group of symmetries of a square sometimes written $D_8$) Apr 20 '18 at 1:46
• I don't understand how that would help me. Apr 20 '18 at 2:06
• I'm trying to say that $J$ is isomorphic to $D_4 = \langle \sigma, \tau \sigma^4 = \tau^2 = (\sigma \tau)^2=1 \rangle$ I think you can show, mapping $(1234) \to \sigma$ and $(12)(34) \to \tau$ that these are isomorphic. What do you know then about the orders of isomorphic groups? Apr 20 '18 at 2:13

Let $H$ be the subgroup generated by $r =(1234)$ and $K$ be the subgroup generated by $s = (12)(34)$. Since $srs^{-1} = (2143) = r^{-1}$, $K$ lies in the normalizer $N_G(H)$. Thus $HK$ is a subgroup of order $\frac{|H||K|}{|H \cap K|} = 8$.