Q: Every group of order 4 is isomorphic to $\mathbb{Z}_{4}$ or the Klein group I wanted to prove that every group or order $4$ is isomorphic to  $\mathbb{Z}_{4}$ or to the Klein group. I also wanted to prove that every group of order $6$ is isomorphic to $\mathbb{Z}_{6}$ or $S_{3}$.

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*For the first one I tried to prove that $H$ (a random group of order 4) is cyclic or the Klein group, because if $H$ is cyclic I can prove that a cyclic group of order $n$ is isomorphic to $\mathbb{Z}_{n}$. Because $H$ has order $4$ it's only possible for elements in $H$ to have order $1$, $2$, $4$ (Lagrange). Say that $H$ is not cyclic. Then all the elements need to have order $1$ or $2$. Not all the elements can have order $1$ so there must be one element of order $2$. Say that $b$ is an element with order $2$. Then take $c$ an element not the unit element or $b$. Then $H=\{e, b, c, bc \}$, so $c$ must have order $2$ because otherwise $H$ would have an order bigger than $4$. This is the Klein group.


*I wanted to do the second one analogously but I can't make a proper proof out of it.
Can someone help and correct me? (I'm so sorry for my English mistakes but I'm really trying.)

A: As @rain1 pointed out, we have a group $G=\{1,a,b,ab\}$, where $a$ and $b$ are different, commute and are not equal to $1$. Let us call $ab=ba=c$. Observe that $a \neq c$ and $b \neq c$. Now look at $a^2$. Then $a^2 \notin \{a,c\}$, so either $a^2=1$ or $a^2=b$. Symmetrically, either $b^2=1$ or $b^2=a$. So there are $4$ cases to consider, but by symmetry in $a$ and $b$ this boils down to only $2$. Firstly, $a^2=1$ and $b^2=1$, in this case $G \cong V_4$. And secondly, if $a^2=1$ and $b^2=a$, then $b^4=1$ and $G \cong C_4$. So no need of the structure theorem of abelian groups.
For groups of order $6$ you can proceed in a similar, but slightly more complicated way. Just applying elementary means. No Lagrange, no Cauchy.

A: $\bullet$ Let $G$ be a group of order $4$. By Lagrange's theorem, $o(g)=\{1,2,4\}$ for each $g \in G$. Note that $o(g)=1$ if and only if $g=e$.
If $o(g)=4$ for at least one $g \in G$, then $G$ is cyclic. Hence $G \cong {\mathbb Z}_4$.
Otherwise, $o(g)=2$ for each $g \in G$, $g \ne e$. So if $G=\{e,a,b,c\}$, then $a^2=b^2=c^2=e$. Now $c=ab$ (since $c \ne a,b,e$ and $c \in G$), the correspondence $a \leftrightarrow (1,0)$, $b \leftrightarrow (0,1)$, $c \leftrightarrow (1,1)$ sets up the isomorphism between $G$ and ${\mathbb Z}_2 \oplus {\mathbb Z}_2$. This is the Klein $4$-group.
$\bullet$ Let $G$ be a group of order $6$. By Lagrange's theorem, $o(g)=\{1,2,3,6\}$ for each $g \in G$. Note that $o(g)=1$ if and only if $g=e$.
If $o(g)=6$ for at least one $g \in G$, then $G$ is cyclic. Hence $G \cong {\mathbb Z}_6$.
Otherwise, $o(g)=2$ or $3$ for each $g \in G$, $g \ne e$. Assuming Cauchy's theorem, there exist elements of order $2$ and $3$, say, $a$ and $b$, respectively. This already accounts for distinct elements $e,a,b,b^2$ in $G$. In addition, we must also have $ab,ab^2,ba,b^2a \in G$.
We can show that none of these four elements can equal $e,a,b,b^2$ by eliminating each case.
We can also eliminate $ab=ab^2,ba$, leaving us with the possibility $ab=b^2a=b^{-1}a$. We can similarly eliminate $ba=b^2a,ba$, leaving us with the possibility $ba=ab^2=ab^{-1}$. Now both these possibilities must happen in order that only two of these elements are distinct.
This leaves us with $G=\{e,a,b,b^2,ab,ba\}$, with $ab^2=ba$ and $b^2a=ab$. The correspondence $a \leftrightarrow (1\:2)$ and $b \leftrightarrow (1\:2\:3)$ sets up an isomorphism between $G$ and $S_3$. $\blacksquare$
I have tried not to use the concept of normality since only basic tools are meant to be used. That every group of even order has an element of order $2$ can be readily proved by a parity argument. For the corresponding result for groups of order multiple of $3$, I have resorted to Cauchy's theorem. Without this, we would need to show that not every element can be of order $2$, and a little more would need to be written.

A: Let $G$ be a group of order 4. Assume for contradiction that it is not abelian, i.e. we have elements $a,b$ that do not commute: then $1, a, b, ab, ba$ are 5 distinct elements.
Therefore $G$ is abelian, and by the structure theorem for abelian groups it must be isomorphic to one of $C_4$ or $C_2 \times C_2$.
