2

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

Jack
  • 625
  • 1
    What does the notation $G = { (1, 2, 3, ... n)^a\cdot(n+1, n+2, n+3, ... 2n)^b \mid 0 \leq a,b \leq n-1 }$ mean? – DHMO Apr 17 '17 at 15:57
  • Sorry maybe I should have been more specific. $(1, 2, 3, ... n)$ and $(n+1, n+2, n+3... 2n)$ represent permutations (in k-cycle notation) - and I'm examining the group formed by taking products of powers of these cycles. – Jack Apr 17 '17 at 16:00
  • @Jack no power of (1,2,...n) belongs to the subgroup generated by (n+1,...,2n) – i. m. soloveichik Apr 17 '17 at 16:04

3 Answers3

4

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 $Z_n \times Z_n \not\cong Z_{n^2}$.

Trevor Gunn
  • 27,041
1

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

lhf
  • 216,483
  • I'm not sure whether that's correct - for instance the identity is an element of the group but it's order is 1 - not n? – Jack Apr 17 '17 at 16:08
  • I said, at most $n$. – lhf Apr 17 '17 at 16:08
  • Thanks for your reply Ihf - and sorry for not catching on - but can you explain why $g^n$ must equal 1 and, why that means the order of every element in G is at most $n$. – Jack Apr 17 '17 at 16:16
1

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.

Daniel Donnelly
  • 21,342
  • Thank you for your reply. However, unfortunately, I haven't seen much of the notation which you've used before. I have neither encountered "direct products" of groups. I don't suppose there is a simpler answer akin to Ihf's? – Jack Apr 17 '17 at 16:22
  • @Jack what's an lhf? – Daniel Donnelly Apr 21 '17 at 04:23