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.
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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.)
Best Answer
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.