Let $G_1 = <(123)(45)>$ and $G_2 = \{e, (12)(34), (567), (576), (12)(34)(567), (12)(34)(576)\}$. I want to find an isomorphism between $G_1$ and $G_2$. I already know that:
(1 2 3)(4 5) $\rightarrow$ (1 2)(3 4)(5 6 7)
(1 3 2) $\rightarrow$ (5 7 6)
(4 5) $\rightarrow$ (1 2)(3 4)
(1 2 3) $\rightarrow$ (1 2)(3 4)(5 6 7)
(1 3 2)(4 5) $\rightarrow$ (5 7 6)
e $\rightarrow$ e
I usually find isomorphism between two groups by creating multiplication tables for each group and finding the patterns, but apparently this process is tedious and time-consuming if there are many elements. Is there an efficient way to find isomorphism in this case?
Let $a:=(1\,2\,3)(4\,5)$, then $G_1=\{e,a,a^2,a^3,a^4,a^5\}\ $ and $a^6=e$.
Let $b:=(1\,2)(3\,4)(5\,6\,7)$, then - as you calculated - we have $$b^2=(5\,7\,6),\ \ b^3=(1\,2)(3\,4),\ \ b^4=(5\,6\,7),\ \ b^5=b^{-1}=(1\,2)(3\,4)(5\,7\,6)$$ So that $\langle a\rangle$ and $\langle b\rangle$ are both cyclic subgroups of order $6$, and sending the generator to generator, namely $a\mapsto b$, will uniquely extend to a homomorphism, as it necessarily satisfies $a^k\mapsto b^k$ for every $k\in\Bbb Z$, and this homomorphism is clearly bijective, hence is an isomorphism.