Prove that if a group $G$ has $|G|$ = 6 then G is isomorphic to either $\mathbb{Z}_6$ or $S_3$.
I have the proof by contradiction but I was wondering if there was a direct proof instead in case I come across a situation where the size is considerably bigger.
- Here is the proof by contradiction:
Assume otherwise. If $|G|$ = 6 and no element of order 6, then all elements of G are order 1 (the identity), order 2, or order 3.
If all elements of order 2, then choose $x, y \in G$, not equal to each other $x*y$ not equal to 1, $x$, or $y$ either so
$\{1,x,y,xy\}$ is a subgroup (isomorphic to $V_4$). But size 4 but contradicts Lagrange (4 does not divide $|G|$=6).
So $G$ must have an element of order 3. Call this $z$. Then $H = \{1,z,z^2\}$, is a subgroup of order 3, i.e. of index 2, hence is normal.
Let $w$ be outside of $H$. Then $G = \{1,z,z^2, w, w*z, w*z^2\}$ by using cosets and further $wH * wH = H$ so $w^2$ in $H$.
If $w^2 = z$ then $w$ would be of order 6. Similarly if $w^2 = z^2$. Thus we must have $w^2 = 1$.
Also $wH = Hw$ so $w*z = w, z*w$ or $z^2*w$. Can see $w*z = w$ not possible since $z$ is not the identity.
If $w*z = z*w$, then $G$ is abelian, but then $(w*z)^2 = w^2*z^2 = z^2$ and $(w*z)^3 = w^3*z^3 = w^3 = w$, and so $w*z$ is order 6 in that case.
Thus we must have $w$ of order 2, $z$ of order 3, and $w*z= z^2*w = z^{-1}*w$.
This is the same relations as $S_3$ with $z = (1\ 2\ 3)$ and $w = (1\ 2)$ or equivalently $D_3$ with $z$ = rotation, $w$ = reflection.
It looks good, for the most part. The only piece I see that wants more justification is your claim that $\{1,x,y,xy\}$ is a subgroup of $G$ if all elements of $G$ have order $2$. For example, it isn't immediately apparent what $(xy)x$ should be. Still, you should easily be able to show that it is $y,$ whence $xy=yx,$ and so you reach the contradiction as before.
Added: Incidentally, what you have is, in fact, a direct proof! Given any two statements $p$ and $q,$ the statement "$p$ or $q$" is equivalent to the statement "if not $p,$ then $q.$" In this case, you wish to show that $G$ is isomorphic to $\Bbb Z_6$ or $G$ is isomorphic to $S_3.$ Equivalently, you wish to show that if $G$ is not isomorphic to $\Bbb Z_6,$ then $G$ is isomorphic to $S_3;$ this is exactly what your proof accomplishes!