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Sorry for posting a second question on this topic, abstract algebra is taking a bit longer to get my head around.

I'm trying to work out if this is cyclic or not, and find all generators or show no generator exists.

$\langle(12)(34)(56),(145)(236)\rangle \leq S_6$

Just in case there's differences in notation, that's just the subgroup generated by the elements $(12)(13)(56)$ and $(145)(236)$ of $S_6$. The problem is, my notes only show one example like this, but both permutations were even, and it was in $S_4$ so they just showed that you could generate $A_4$ (by literally showing you could make every even permutation) using the two elements. Given how one of the elements here is odd, and we're dealing with $S_6$ it doesn't look like a good method.

So far, I've noted that, letting $\sigma = (12)(34)(56)$ and $\tau = (145)(236)$ we get that $o(\sigma) = 2$ and $o(\tau) = 3$

Also, that $\sigma \tau = (135246) = \tau \sigma$, $(\sigma \tau)^2 = (\tau \sigma)^2 = (154)(326) = \tau^{-1}$

Lastly that $(\sigma \tau)^3 = (\tau \sigma)^3 = (12)(34)(56) = \sigma = \sigma^{-1}$

I get a bit stuck as to where to go at this point, although I can't see how it's even possible this is cyclic. Thanks again, this place has already been a massive help!

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you have already showed that the element $\tau \sigma$ generates the group $\langle(12)(34)(56),(145)(236)\rangle$, therefore it is a cyclic group in particular $C_6$ – user58512 Mar 15 '13 at 15:43
up vote 4 down vote accepted

The most important observation is that $\sigma$ and $\tau$ commute; without that you could never have that they generate a cyclic subgroup. But once this is checked, you know they generate a commutative subgroup; you can ignore the rest of $S_6$ and apply the theory of Abelian groups. In an Abelian group, two elements $a,b$ with relatively prime orders $p,q$ always have the property that $ab$ has order $pq$: the relative primality ensures that $\langle a\rangle\cap\langle b\rangle=\{e\}$, so $(ab)^i=a^ib^i=e$ implies $a^i=e=b^i$, hence $\operatorname{lcm}(p,q)=pq\mid i$. In your case you have elements $\sigma,\tau$ of relatively prime orders $2,3$, so $\sigma\tau$ has order $6$ and generates $\langle\sigma,\tau\rangle$, which is therefore cyclic.

The argument of course shows that whenever in any group you have commuting elements of relatively prime orders, they generate a cyclic group (you can even have more than two generators to start with, provided their orders are pairwise relatively prime; just iterate the argument).

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Oh, good show! Excellent answer! – user1729 Mar 15 '13 at 16:12
@Marc Thank you for the excellent explanation! – Noble. Mar 15 '13 at 18:06

It's obvious by inspection that $(12)(34)(56)$ has order 2 and $(145)(236)$ has order 3. Then you can put them together to get an element of order 6: $(12)(34)(56)\cdot(145)(236) = (135246)$ and now you can check that $(135246)$ generates $\langle (12)(34)(56), (145)(236) \rangle$ by computing $(135246)^3$ and $(135246)^4$.

This shows that every element of the group $\langle (12)(34)(56), (145)(236) \rangle$ may be written as $(135246)^i$ for some $0 \le i < 6$ so the group is $C_6$.

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Thank you for the answer, I understand it now! – Noble. Mar 15 '13 at 18:05

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