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Find a subgroup $H$ of $S_6$ such that $D_3$ is isomorphic to $H$

I really need help in this question perparing for a test.

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Another approach to this problem is to note that $D_3$ is a group of six elements. Therefore the group action of $D_3$ on itself by (say) left multiplication is a mapping of $D_3$ to permutations of six things $S_6$.

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If you know that in fact $\,D_3\cong S_3\,$ then all you need to do is to choose a subgroup of $\,S_6\,$ that only "moves" three elements.

For example, taking $\,S_n\,$ as the group of permutations on $\,\{1,2,3,...,n\}\,$, you can take $$H:=\{\sigma\in S_6\;:\;\sigma(i)=i\,\,\,\forall \,i=4,5,6\}$$

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Hint: $D_3$ is generated by an element of order $3$ and an element of order $2$. Try finding an element of order $3$ and an element of order $2$ in $S_6$, and let $H$ be the subgroup generated by them.

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The cycle $(1~2~3)$ has order $3$ and the transposition $(4~5)$ has order $2$. They generate $H$ of order $6$, but it is not isomorphic to $D_3$. In fact many choices will not work; for $ (1~2~3)$ and $(1~4)(2~5)(3~6)$ the subgroup has order $18$. – Marc van Leeuwen Aug 2 '12 at 6:33

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