# Prove an abelian group of order six is cyclic

Suppose$G$ is an abelian group and $|G|=6$, prove that $G=\{e,g,g^2,...,g^5\}$ for some $g$ with $g^6=e$.

My attempt:

1. If $G$ consists only of elements of order 2, then $|G|=2^m$ for some $m$. $6 \neq 2^m \Rightarrow$ $G$ has at least one element $g'$ with $|g'| \neq 2$
2. By application of Lagrange's theorem(considering subgroups of elements), all elements of $G$ should have orders 1,2,3 or 6, thus $G$ must contain an element of order 3. Call it $b$.
3. Since $|G|$ is even, it contains an element of order 2, call it $a$.

Now I can say $G=\{e,a,b,b^2,ab,ab^2\}$

But how should I prove $G$ is cyclic?

• In any Abelian group, if the orders ord(x) and ord(y) exist and are co-prime then ord(xy)=ord(x).ord(y). Commented Nov 3, 2015 at 11:28
• Why is order of group of the form $2^m$ in the first point. Commented Sep 27, 2022 at 15:22

Just compute the powers of $ab$ (using the relations $a^2=b^3=1$ on the way).
• And you can stop when $(ab)^3\ne e$...
Since $ab\ne e$, $(ab)^2\ne e$, $(ab)^3\ne e$, it must have order $6$.
\begin{aligned}(a b)^2=a^2b^2=b^2&\implies (a b)^3=(a b)(a b)^2=(a b)b^2=a b^3=a\\ &\implies (a b)^4=(a b)(a b)^3=(a b)a=a^2b=b\\ &\implies (a b)^5=(a b)(a b)^4=(a b )b=a b^2.\end{aligned}