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Let $G$ be a subgroup of the symmetric group $S_6$ given by $G=\langle(123),(456),(23)(56)\rangle$. Show that $G$ has four normal subgroups of order 3.

I may be missing something, but I can only find two of them: $\langle(123)\rangle$ and $\langle (456)\rangle$. Suppose there is another one. It must be singly-generated, either by an element of the form $(abc)$ or $(abc)(def)$. The subgroup must be closed under conjugation with $(123),(456)$, etc. I always end up with more than three elements.

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Hint: $(123)$ and $(456)$ commute, so $(123)(456)$ has order $3$.

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  • $\begingroup$ Silly me. I thought that conjugations with $(123),(132),(456),(465)$ result in more than 3 elements. Thanks. $\endgroup$ – abc Jun 20 '13 at 8:13
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You are on the right lines. However as Seirios has just said, $(123)(456)$ has order $3$ and clearly is normalised by your subgroup. Thus $\langle (123)(456)\rangle$ will be another normal subgroup of order $3$.

However, for exactly the same reasons you can take $(123)^2(456)=(132)(456)$ and then this will also give you an element of order $3$ that generates the normal subgroup $\langle (132)(456)\rangle$.

Note of course that we could have done the same thing with $(123)(456)^2=(123)(465)$, but then $\langle (123)(465)\rangle=\langle (132)(456)\rangle$ since $(123)(465)=\left((132)(456)\right)^2$.

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