Proving that a Group isn't Cyclic

Question

I'm trying to prove that the group below isn't cyclic:

$G = \{ (1, 2, 3, ... n)^a\cdot(n+1, n+2, n+3, ... 2n)^b \mid 0 \leq a,b \leq n-1 \}$

To do this, I'm trying to show that none of the elements of $G$ have order $n^2$ - however I've made little progress.

Any help would be much appreciated,

Jack

Answer 1

Let $\alpha = (1,2,3,\dots,n)$ and $\beta = (n + 1, n + 2, \dots, 2n)$. Then given any element $\alpha^a\beta^b$, $(\alpha^a\beta^b)^n = \alpha^{na}\beta^{nb} = e$ since $\alpha$ and $\beta$ commute. You might also like to show that $G \cong Z_n \times Z_n$ where $Z_n$ is the cyclic group of order $n$ and the isomorphism is $\alpha^a\beta^b \mapsto (a,b)$. Then you would need to show that $Z_n \times Z_n \not\cong Z_{n^2}$.

Answer 2

Indeed, if $g \in G$, then $g^n=1$. So the order of every element in $G$ is at most $n$.

Answer 3

The group is generated by two cycles $\sigma = (1, \dots, n), \tau = (n+1, \dots, 2n)$ in $S_{2n}$. These cycles are disjoint obviously.

Define $\Bbb{Z}_n \times \Bbb{Z}_n \xrightarrow{\phi} S_n$ by $a \times b \mapsto \sigma^a \tau^b$. Then this is an embedding as $\phi(a,b) = \text{id} \implies a = b = 0$ or else $\sigma^a \tau^b = \text{id}$ so that $\tau$ is not only not a disjoint cycle from $\sigma$ it is an iterate of it ($\sigma^k$, for some $k$).

But according to this lemma: The direct product of two cyclic groups is cyclic iff their orders are coprime.

It's nice to use other results. It reduces the workload of what you're trying to prove.