Cyclic subgroups and generators

Solution 1

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).

Solution 2

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 $C_6$.

Author by

Noble.

Updated on August 01, 2022