I'm trying to prove the following assertions.
Classify groups of order less than or equal to $4$ along the following lines:
(a) There is one group of order $2$ up to isomorphism.
(b) There is one group of order $3$ up to isomorphism, which is isomorphic to $\mathbb{Z}/3\mathbb{Z}$.
(c) There are two groups of order $4$ up to isomorphism, one isomorphic to $\mathbb{Z}/4\mathbb{Z}$ and one isomorphic to $\mathbb{Z}/2 \times \mathbb{Z}/2$.
I'll try to prove each of these in turn.
Groups of Order Two.
This follows by writing the multiplication table. We call these elements $\{e, a\}$, where $e$ is the identity element. By definition, we have $e \cdot e = e$ and $e \cdot a = a \cdot e = a$. But $a$ needs to have an inverse and the only output we haven't defined is $a \cdot a$, so we must have $a \cdot a = e$.
While I understand why this implies that all groups of order $2$ are "secretly the same group," I don't see how this gives rise to an isomorphism, even though it's obvious to me that I can map identity element to identity element and non-identity element to non-identity element, and only one possible isomorphism exists. Is this enough to prove that there is only one group, or is there additional explanation I'm missing?
Groups of order $3$.
Let $H$ be a group of order $3$. By Lagrange's theorem, elements of $H$ either have order $1$ or order $3$ since $3$ is prime, and the order of an element of $H$ must divide $|H| = 3$. The only element of order $1$ is $e$ since the identity element is unique, so any remaining element must have order $3$. Let $H = \{e, a,b\}$. Then $a^3 = 1$ and $b^3 = 1$. In particular, $a^2 \neq 1$ and $b^2 \neq 1$ As $a$ and $b$ must be invertible, we must have $ab = ba = 1$. This completely defines the multiplication table for $H$. Furthermore, as $a$ and $b$ have order $3$, $\langle a \rangle$ and $\langle b \rangle$ have order $3$, so $H$ is cyclic, generated by either $a$ or $b$. Any finite cyclic group of size $n$ is isomorphic to $\mathbb{Z}/n$ via the bijection $\mathbb{Z}/n \to H$ sending $k \mapsto a^k$.
I run into the same issue where I can prove that there is only one multiplication table but I don't understand why this implies the definition of isomorphism, i.e., why there is a bijective homomorphism.
Groups of Order Four
This is the part of the problem I've been having the most trouble with.
Let $H$ be a group of order $4$. By Lagrange's theorem, the elements of $H$ either have orders $1$, $2$, or $4$. Suppose $H$ contains an element of order $4$. Call this element $a$. Then $\langle a \rangle \lhd H$ and $|\langle a \rangle| = |H|$, so we have $H = \langle a \rangle$. As $H$ is cyclic of order $4$, we have $H \cong \mathbb{Z}/4\mathbb{Z}$. Suppose $H$ lacks an element of order $4$. Then the three non-identity elements of $H$ have order $2$. Write $H = \{e, a, b, c\}$ where $e$ is the identity element and $a$, $b$, and $c$ have order $2$. Then $a^2 = b^2 = c^2 = e$. We need only define the products $ab$, $ba$, $ac$, $ca$, $bc$, and $cb$. Notice that $ab \neq a$ because, otherwise, $b = e$. Similarly, $ab \neq b$, $ba \neq a,b$, $ac \neq a,c$, $ca \neq a,c$, $bc \neq b,c$, and $cb \neq b,c$. Therefore, $ab = ba = c$, $bc = cb = a$, and $ac = ca = b$.
I know at this point that I've defined the structure of the Klein four-group, but I don't know how to write down the isomorphism. If I had a definition in mind, I suppose I could say this has the same multiplication table as the Klein four-group and is therefore isomorphic to it. Because I've defined a single multiplication table for a group of order $4$ without an element of order $4$, there are only two groups.
I would appreciate some feedback on these proofs. I'm in particular interested in the connection between the multiplication table being "the same" and an isomorphism and in how to write down an isomorphism with the klein four-group.