This is section 2.4 question 7 in Dummit and Foote.
Show that the subgroup of $S_4$, $A=\langle (12),(13)(24)\rangle$ is isomorphic to dihedral group of order $8$
Here $D_{4}=\langle r,s:r^4=s^2=e, srs=r^{-1}\rangle$.
I believe I want to use that $(12)(13)(24)=(1423)$ is the element mapped to by $r\in D_4$
But do I know that $\langle (12), (1423)\rangle=\langle (12), (13)(24)\rangle$?
And if I define $\phi:D_4\to S_4$ by $\phi(r)=(1432)$ and $\phi(s)=(12)$.
I can show that $\phi(s)\phi(r)\phi(s)=\phi(r)^{-1}$, since $(12)(1432)(12)=(1324)$
I believe this means that $\phi$ is a homomorphism, In Dummit and Foote they haven't proven this yet though, I'm not sure if there is some other way to prove that this $A\cong D_4$?
Also is proving a bijection enough to show that the map $\psi((12))=s, \psi((1432))=r$, is an inverse? Or do I have to show they have the same order and then show explicitly write a map for each element?
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