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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 $G$?

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    $\begingroup$ By definition, the notation $\langle S \rangle$ denotes the subgroup generated by the set $S$ - ie. the smallest subgroup containing $S$. $\endgroup$ – Prahlad Vaidyanathan Mar 3 '15 at 10:18
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    $\begingroup$ If you write $< (123), (456) >$, then it's the group generated by these two elements. $\endgroup$ – Tlön Uqbar Orbis Tertius Mar 3 '15 at 10:19
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    $\begingroup$ 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 $H$. $\endgroup$ – frabala Mar 3 '15 at 10:20
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    $\begingroup$ @frabala $\langle (456)^2 \rangle = \langle (456) \rangle$ $\endgroup$ – Derek Holt Mar 3 '15 at 11:48
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    $\begingroup$ @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? $\endgroup$ – frabala Mar 3 '15 at 15:14
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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.

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    $\begingroup$ The question asks you to show the subgroups are normal in $G$, not in $H$, and $G$ isn't abelian. $\endgroup$ – Gerry Myerson Mar 3 '15 at 11:58
  • $\begingroup$ The grey part is quite ambiguous, not specifying where the subgroups should be normal. You should never write like this, normality is defined only in terms of some other (parent) group. The next line clears the confusion. $\endgroup$ – Leppala Mar 3 '15 at 14:05

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