# Subgroup of $G=\langle(123),(456),(23)(56)\rangle$

Let $G=\langle(123),(456),(23)(56)\rangle$ be a subgroup of order $18$ of $S_6$.

Clearly $H=\langle(123),(456)\rangle$ is a subgroup isomorphic to $C_3 \times C_3$.

SOLVED: Not sure how $H$ is a subgroup clearly. Could someone please explain how this is so obvious?

Apparently $H$ has subgroups, $A_1=\langle(123)\rangle,A_2=\langle(456)\rangle,A_3=\langle(123)(456)\rangle,A_4=\langle(123)(465)\rangle$.

How is $A_4$ a subgroup?

...these are all normal subgroups

Again how is it obvious that the $A_i$'s are normal in $G$?

• By definition, the notation $\langle S \rangle$ denotes the subgroup generated by the set $S$ - ie. the smallest subgroup containing $S$. – Prahlad Vaidyanathan Mar 3 '15 at 10:18
• If you write $< (123), (456) >$, then it's the group generated by these two elements. – Tlön Uqbar Orbis Tertius Mar 3 '15 at 10:19
• Not sure for this, but I'll give it a try. Elements $(123), (456)$ and $(23)(56)$ are those that generate $G$. So, if you take a subset of the set of generators, then this subset generates a subgroup, in this case $H$. – frabala Mar 3 '15 at 10:20
• @frabala $\langle (456)^2 \rangle = \langle (456) \rangle$ – Derek Holt Mar 3 '15 at 11:48
• @DerekHolt Oh, I mixed up the fives and the sixes there... :P. What I meant is that $(465)^2 = (456)$, so $\langle (465) \rangle$ is a subgroup of $\langle (456) \rangle$, specifically it has two elements only: the generator and the identity. By $(456)$ we mean the identity element, right? – frabala Mar 3 '15 at 15:14

Your $H$ is the subgroup generated by $(123)$ and $(456)$, in the sense that your notation $\langle \cdot \rangle$ explicitly means the subgroup generated by $\cdot$. So it's a subgroup, very clearly.
Since they're disjoint 3-cycles, you also get that it's isomorphic to $C_3 \times C_3$. This is also how you know, for instance, that $A_4$ is a subgroup. It's the subgroup generated by $(123)(465)$. To see that it's a subgroup of $H$, you just need to verity that $(123)(465) \in H$ (which it is - but if you haven't, you should verify it!).
There are very many reasons why the $A_i$ are normal. One way to see is to explicitly check. This set of computations might give you a better understanding of what's going on, so perhaps it's even a good idea.
A different way is to note that $H$ is abelian, and any subgroup of an abelian group is abelian.
Yet another way is to note that these are subgroups of minimal prime index in $H$, and subgroups of minimal prime index are normal. To be fair, I do not think you should know this statement yet. But somehow it's good to get an idea of some of the general statements out there, to see how everything connects.
• The question asks you to show the subgroups are normal in $G$, not in $H$, and $G$ isn't abelian. – Gerry Myerson Mar 3 '15 at 11:58