# Does $S_3 \times \mathbb Z_6$ have a subgroup of order 9.

Does $$S_3 \times\mathbb Z_6$$ have a subgroup of order $$9$$? where $$S_3$$ is the symetric group. Clearly $$|S_3| \times |\mathbb Z_6|=36$$ and $$9$$ does divide $$36$$, so its possible. But i can not figure out why or why not there would be a subgroup of order $$9$$.

• Does $S_3$ have a subgroup of order $3$? Does $Z_6$ have a subgroup of order $3$? – Lord Shark the Unknown Jun 1 at 14:29

## 2 Answers

Hint: Both $$S_3$$ and $$\mathbb Z_6$$ have subgroups of order $$3$$.

If you know about Sylow Subgroups and the existence of these. Then, a 3-Sylow Subgroup of a group of order $$36=3^2\times 2^2$$ will have order $$3^2=9$$.