If $G$ has order $4$ and has an element of order $4$, then $G$ is isomorphic to $\mathbb{Z_4}$.
Can someone briefly explain why this is true? I understand that $|G| = 4$, but I don't understand why, if one element has order $4$, then the two sets are necessarily isomorphic.
On
If you try to write out a multiplication table for a group with 4 elements (lets call it $G = \{e, a, b, c\} $) then you will see that there are only two possible multiplication tables. So, up to isomorphism there are only two different groups of order $4$. One of these has an element of order $4$, the other doesnt. Now its left to you to check if the group with the element of order $4$ is isomorphic to $\mathbb{Z}_4$
On
Note that $\mathbb{Z}/n\mathbb{Z}$ is generated by $1$. If you have a group $G$ of order $n$ with an element $g$ of order $n$, then $g$ is a generator for the group. So, the natural map to consider is just $g\leadsto 1$. Note that the only difference here is that you will be working with exponents in $G$ and addition in $\mathbb{Z}/n\mathbb{Z}$. Note that since each member of $G$ is of the form $g^m$ for some $m\in \mathbb{Z}$, the homomorphism $G\rightarrow \mathbb{Z}/n\mathbb{Z}$ is completely determined by $g\leadsto 1$, so you just have to check that it's 1-1 and onto, which is easy.
On
By definition, a group $G$ is cyclic if it contains an element whose powers generate $G$, i.e. if $\exists x \in G$ such that $\{x^n: n \in \mathbb{Z} \} = G$. Since $G$ has order 4 and contains an element $a$ of order 4, the elements $a,a^2,a^3$ and $1$ are distinct and exhaust all the elements of $G$. Thus, $G$ is the cyclic group $\langle a \rangle$. Any two cyclic groups of the same order are isomorphic: for eg, $G=\langle a \rangle$ and $\mathbb{Z}_4 = \langle 1 \rangle$ are isomorphic via the map $a^k \mapsto k$.
Let me expound upon Krjin's answer. Your group $G$ is of the form $\{e,a,b,c\}$. Suppose $a$ is the element of order $4$. Then there are a couple of options: $a^2 = b$ or $a^2 = c$. Without loss of generality, assume $a^2 = b$, then $b^2 = e$.
From here, $ab = a^3$. This cannot be $a$ or $b$ either and so $ab = c$. Hence $c^2 = a^6 \neq e$ and $c^3 = a^9 \neq e$, giving $c^4 = e$. In recap:
In $\Bbb Z_4$, there are two elements of order $4$ and one element of order $2$ and your group $G$ is cyclic with these same properties. Do you see how to proceed?