# Cyclic subgroups and generators

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!

• 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

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).

• 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
• how can you tell if they commute or not straight away though? Is there a trick to use or is it quicker just to verify by hand? – JustWandering Apr 5 '19 at 16:57
• @JustWandering: There is a quick way to conjugate one permutation (written in cycle form) by another, namely apply the other permutation to the elements in the cycles. If that gives an equivalent cycle form, the conjugation changed nothing and the two commute, which is what happens in the example. – Marc van Leeuwen Apr 5 '19 at 20:22

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 $C_6$.

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