I am trying to show that $\mathbb{Z}/4$ with addition is isomorphic to the non-zero elements of $\mathbb{Z}/5$ with multiplication.
After writing out the multiplication tables, I've found that the elements of $\mathbb{Z}/4\mathbb{Z}$ have the following orders. (I'm not sure if it's standard to use vertical bars or brackets, but I'm going to use brackets and subscripts to distinguish the two groups.) \begin{align*} |[0]_4| &= 1 \\ |[1]_4| &= 4 \\ |[2]_4| &= 2 \\ |[3]_4| &= 4. \end{align*} As for $\mathbb{Z}/5 - \{0\}$: \begin{align*} |[1]_5| &= 1 \\ |[2]_5| &= 4 \\ |[3]_5| &= 4 \\ |[4]_5| &= 2. \end{align*} An isomorphism has to preserve orders, so I need to map: \begin{align*} [0]_4 &\mapsto [1]_5 \\ [2]_4 &\mapsto [4]_5. \end{align*} At this point, I believe I have a choice for the elements of order $4$. I know I need inverses to preserve inverses and to preserve the multiplication table, so it's possible that only one works. (Or is it sufficient to preserve orders? I'd appreciate some help with the intuition.) From the multiple tables, it was clearer to me to send: \begin{align*} [1]_4 &\mapsto [3]_5 \\ [3]_4 &\mapsto [3]_5. \end{align*} The map that accomplishes this is $f: \mathbb{Z}/4\mathbb{Z} \to (\mathbb{Z}/5\mathbb{Z})^{\times}$ defined by $$ f([x]_4) = [2]_5^{[x]_5}. $$ I need to show four things:
This map is well-defined: if $[x]_4 = [y]_4$, then $[2]_5^{[x]_5} = [2]_5^{[x]_5}$.
This map is a homomorphism: for every $[x]_4, [y]_4 \in \mathbb{Z}/4$, $$ f([x]_4 [y]_4) = f([x]_4)f([y]_4). $$
This map is injective: either (directly) $f([x]_4) = f([y]_4)$ implies $[x]_4 = [y]_4$ or $\mathrm{ker}(f)$ is trivial.
This map is surjective (though this is implied from it being an injective homomorphism between groups of equal order): for every $[y]_5 \in (\mathbb{Z}/5)^{\times})$, there exists $[x]_4 \in \mathbb{Z}/4$ such that $f([x]_4) = [2]_5^{[x]_5} = [y]_5$.
I'm ok with proving the homomorphism property. For $[x]_4, [y]_4 \in \mathbb{Z}/4$, the exponent properties and commutativity of $(\mathbb{Z}/5\mathbb{Z})^{\times}$ imply \begin{align*} f([x]_4 [y]_4) &= f([xy]_4) \\ &= [2]_5^{[xy]_5} \\ &= [2]_5^{[x]_5 [y]_5} \\ &= [2]_5^{[x]_5} [2]_5^{[y]_5} \\ &= f([x]_4) f([y]_4). \end{align*} I'm struggling to prove well-definedness. For injectivity, I'd like to invoke injectivity of the exponential function $2^x$, which requires that I invoke logarithms mod $2$, but I'm not sure if I'm allowed to take a logarithm in mod $5$. I'm also not sure how to prove surjectivity, but injectivity, I believe, suffices because the groups have the same order.
I'd appreciate some direction or hints on how to prove well-definedness and bijectivity.
You're overthinking.
Let $G=\Bbb Z_4, H=\Bbb Z_5^*$.
There are four elements in $H$ by inspection and
$$\begin{align} [2]_5^0&=[1]_5,\\ [2]_5^1&=[2]_5,\\ [2]_5^2&=[4]_5,\\ [2]_5^3&=[2\times 4]_5=[3]_5,\\ [2]_5^4&=[2\times 3]_5=[1]_5. \end{align}$$
Thus $H=\langle [2]_5\rangle$ is cyclic.
There are only two groups of order four up to isomorphism: $\Bbb Z_2^2$ and $G$.
Hence $G\cong H$.
Isomorphisms of cyclic groups are determined by where the generator is sent; simply:
$$\begin{align} \varphi: H&\to G,\\ [2]_5&\mapsto[1]_4 \end{align}$$
would suffice.