# If $G=\left<(12),(34),(45)\right>\subset S_5$, then $G\cong C_2\times S_3$

Let $G=\left<(12),(34),(45)\right>\subset S_5$. Show that $G\cong C_2\times S_3$. So my first idea was to set $a=(12)$, $b=(34)$ and $c=(45)$ and remark that $$G=\left<a,b,c\mid ab=ba,ac=ca, a^2=b^2=c^2=1, cb=bcbc\right>.$$

Then if $S_3=\left<(12),(123)\right>$ and $C_2=\left<g\mid g^2=1\right>$.

Then, the morphism I have in my mind is \begin{align*} G&\longrightarrow C_2\times S_3\\ 1&\longmapsto (1,1)\\ a&\longmapsto (g,1)\\ b&\longmapsto (1,(12))\\ c&\longmapsto (1,(23)). \end{align*}

What I did is to write completely $G$ as $G=\{1,a,b,b,ab,ac....\}$ and I check by hand that it was a group isomorphism.

The problem : It's very long to do as I did, and I was wondering I there is a shorter method first, or a method that gives you immediately that the homomorphism I wrote is the isomorphism researched.

• I'm not sure this is really shorter, but I would've tried to show that $G$ is the inner direct product $\langle (12) \rangle \times \langle (34),(45)\rangle$. It's pretty obvious the first factor is $C_2$ and the second is $S_3$ then. – sTertooy Jun 21 '16 at 13:24
• @idm: Where is this problem from? – Alphonse Jun 21 '16 at 14:14
• @Alphonse: An exam I had yesterday. – idm Jun 22 '16 at 8:34

Enumerating the elements of $G$ we get $$\{e,(34),(45),(35),(345),(543)\}\cup \{(12),(12)(34),(12)(45),(12)(35),(12)(345),(12)(543)\}$$ Verify that the subgroup generated by $(34)$ and $(45)$ is normal, and the subgroup generated by $(12)$ is normal. Obviously they intersect trivially, and the suggestively written union above shows that they generate the group. By definition, $G$ is the direct product of $\langle (12)\rangle$ and $\langle (34),(45)\rangle$.