Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

3 Answers 3

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 $S_6$.

share|improve this answer

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 $$H:=\{\sigma\in S_6\;:\;\sigma(i)=i\,\,\,\forall \,i=4,5,6\}$$

share|improve this answer

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.