Show that $G=\{0,1,2,3\}$ over addition modulo 4 is isomorphic to $H=\{1,2,3,4\}$ over multiplication modulo 5
My solution was to brute force check validity of $f(a+b)=f(a)f(b)$ for all $a,b\in\{0,1,2,3\}$ where i took $f(x)$ as $f(x)=x+1$. I would like to know if there's a more elegant way?
Both are cyclic groups of order 4. This is enough to say they are isomorphic because all cyclic groups of order $n\in\mathbb{N}$ are isomorphic to $\mathbb{Z}_n$, but the general idea is that you're mapping one cyclic generator to another.
Notice that $1$ generates $G$ and $2$ generates $H$. So now define $\phi:G\rightarrow H$ as $\phi(a)=2^a\pmod5$. Then $\phi(a+b)=2^{a+b}\pmod5=2^a2^b\pmod5=\phi(a)\phi(b)$. Thus, $\phi$ is operation-preserving. I'll leave it up to you to show that $\phi$ is bijective.
In general, you need to prove the existence of a bijective homomorphism between the two groups.
In practice, there is only one cyclic group of each order, $\Bbb Z_n$. Here can use that fact to establish the result.
To wit, $\Bbb Z_p^\times$ is known to be cyclic of order $p-1$.