# Finding what $\langle(135)(246),(12)(34)(56)\rangle\subset S_{6}$ is isomorphic to

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 $\mathbb{Z}_6$.

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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. –  JSchlather Jun 8 '12 at 21:19

$(135)(246)(12)(34)(56)=(145236)$, 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.
Maybe also include the idea of taking exponents mod the order, so that $(ab)^i = a^{i \mod 3} b^{i \mod 2}$ so that $(ab)^3 = a^3 b^3 = a^0 b^1 = b$, $(ab)^4 = a^1 b^0 = a$, and $(ab)^6 = a^0 b^0 = 1$, so that $\langle a,b \rangle$ contains $ab$ of order 6, and $\langle ab \rangle$ contains $a$ and $b$, so $\langle a,b \rangle = \langle ab \rangle$ is cyclic of order 6. In other words, not only permutations, but words too. I liked polynomials better when I first started learning group theory, and manipulating words was closer to what I knew. –  Jack Schmidt Jun 8 '12 at 21:16
I wonder if we can avoide the arguments about $\pi^3$ and $\pi^4$ and get a bit more general approach ? althogh this is a basic question I am fermiler with more advanced claims... –  Belgi Jun 8 '12 at 21:19
@Belgi: your lcm claim is close to correct as long as the elements commute (see Jacob's answer). However if $a$ has order 4, and $b$ has order 4, and $a,b$ commute, then it is hard to say how big $\langle a,b \rangle$ is. It might have any order in {4,8,16}. When the orders are coprime, then you do get a cyclic group, for exactly the $\pi^3$ and $\pi^4$ kinds of reasons. I generally call this the Euclidean algorithm or chinese remainder theorem. –  Jack Schmidt Jun 8 '12 at 21:21