6
$\begingroup$

I am doing an exercise that asks me to find what $\langle(135)(246),(12)(34)(56)\rangle\subset S_{6}$ is isomorphic to.

I am allowed to only use the groups $D_n,S_n,\mathbb{Z}_n$ and the direct sums ) where $S_n$ is the permutatin group of $n$ elements, $D_n$ is the dihedral group of order $2n$.

I have noted that the first element is of order $3$ and that the second one is of order $2$. I also noted that these elements commutes hence generate an abelian group. I can also say that this group is of order at least $6$ since $gcd(2,3)=1$.

How can I find what is Finding $\langle(135)(246),(12)(34)(56)\rangle\subset S_{6}$ ? If there was a good argument that say that this group is at most of order $6$ then I can clain that since the only groups of order $6$ are $S_3$ and $\mathbb{Z}_6$ and $S_3$ is non abelian then this group is isomorphic to $\mathbb{Z}_6$.

Can someone please help with this problem ?

  • 1
    I deleted by answer, because I think Brian's is much better. But I did want to answer your question concerning the order of $gh$. In an abelian group the best you can say is that the order of $gh$ divides the $\mathrm{lcm}(o(g),o(h))$. Notice for instance if $h=g^{-1}$ then $gh$ has order $1$. In the non-abelian case there's nothing meaningful you can say in general. For instance $g$ and $h$ can both have finite order, but $gh$ have infinite order.2012-06-08

2 Answers 2

8

You already did much of the work when you calculated the product of the two elements.

$(135)(246)(12)(34)(56)=(145236)$, which clearly has order $6$. On the other hand, if this permutation is $\pi$, it’s easy to check that $\pi^3=(12)(34)(56)$ and $\pi^4=(135)(246)$, so $\pi$ generates the same subgroup.

  • 0
    @Belgi: fi$x$ed m$y$ comment as soon as he posted. :-)2012-06-08
0

If you think the group only has 6 elements then just multiply every combination together and see if you get 6 elements.