I'm currently in an introductory group theory class, and I'm supposed to classify all groups of order 6 up to isomorphism without use of more advanced concepts like the Sylow theorems or even the Cauchy theorems. Can someone help check my proof? I'm trying to be as thorough as possible, so I apologize if it's a little bit long-winded.
There are only two groups of order 6 up to isomorphism, the cylic group $\langle \mathbb{Z}_6 \ |\ +\rangle $ and the symmetric group $\langle S_3 \ | \ \circ \rangle $. We prove this by proving the 2 following statements:
- If $G$ is abelian, then $G \cong \mathbb{Z}_6$
- If $G$ is nonabelian, then $G \cong S_3$
To prove the first statement, we suppose for sake of contradiction that all nonidentity elements have an order of 2. Since $|G| = 6$, there are 5 nonidentity elements. Choose 2 distinct nonidentity elements, $x,y,$ and consider that necessarily $x^2 = y^2 = e$. We have that $G$ is abelian, so the set $H = \{e, x, y, xy \}$ is a subgroup of $G$. However this contradicts Lagrange's Theorem, because $4 \nmid 6$. Thus there are elements of $G$ which are of order 3 or 6. If $G$ has an element of order 6, then $G$ is cyclic, and $G \cong \mathbb{Z}_6$. If $G$ has an element of order 3, say $z$, then $|xy| = 6$. (is this step ok?) Again, $G$ is cyclic, so $G \cong Z_{6}$. \
For the second statement, we briefly pause to show that $G$, being a group of an even order, must have an element of order 2. This is because all elements with order larger than 2 must have an inverse of the same order, so there are an even number of such elements. But $G$ has an even order, and hence an odd number of nonidentity elements, so there is an element of order 2.
We notice that the nonidentity elements of $G$ must be of order $2, 3 $ or $6$. $G$ cannot have an element of order 6 because $a \in G, |a| = 6 \implies G = \langle a \rangle $ implies $G$ is cyclic and therefore abelian. Furthermore by the logic given above, $G$ cannot have all nonidentity elements of order 2, because $G$ would again be abelian. So then there exists $a \in G, |a| = 3$. Now take another nonidentity element, $b$, such that $b \not \in \langle a \rangle $. Again, we can do this because there are 5 nonidentity elements and $\langle a \rangle $ would be only 2 of them. Consider $|\langle a \rangle \cap \langle b \rangle|$:
- If $|\langle a \rangle \cap \langle b \rangle| = 1$, then the cycles generated by $a$ and $b$ have no overlap except for the identity element.
- $|\langle a \rangle \cap \langle b \rangle| = 2$ is not possible, since if $|b| = 2$ then $b \in \langle a \rangle $, but $b$ was chosen specifically so that $b \not \in \langle a \rangle$. $|b| = 3$ would also not be possible because $|b| =3$ implies $| \langle a \rangle \langle b \rangle | = 9$, which contradicts Lagrange's Theorem.
- $|\langle a \rangle \cap \langle b \rangle| = 3$ is not possible, since it implies $|b| = 3$, and then $| \langle a \rangle \langle b \rangle | = 9$ contradicts Lagrange's Theorem.
Thus $|\langle a \rangle \cap \langle b \rangle| = 1$ and $|b| = 2$. We then have the 6 elements of $G$ as being $\{ e, a, a^2, b, ba, ba^2\} = \langle a \rangle \cup \langle b \rangle $. Now consider $ab$.
- $ab \neq e$, since $b \not \in \langle a \rangle $
- $ab \neq a$, since $b \neq e$
- $ab \neq b$, since $a \neq e$
- $ab \neq a^2$, since $b \neq a$
- $ab \neq ba$, since $G$ is nonabelian
- $ab = ba^2$ is the only remaining possibility.
As a result, $G = \langle a, b \mid a^3 = e, b^2 = e, ab = ba^2 \rangle $. This is the same group presentation as for $S_3$, hence $G \cong S_3$
