# $G$ is a group of order $60$. Will $G$ always contain a subgroup of order $6$?

$$G$$ is a group of order $$60$$. Will there be a subgroup of order $$6$$?

Alternating group $$A_5$$ has a subgroup of order $$6$$. That is the group generated by this set $$\{(123), (23) (45)\}$$.

Will we be able to prove that there always exists a subgroup of order $$6$$ in a group of order $$60$$?

Can anyone help me to understand by giving a hint?

No, that's not always true. Take for example $$G = C_5 \times A_4$$. This group has order $$60$$ and no subgroups of order $$6$$. If you know that $$A_4$$ has no subgroups of order $$6$$ (it is the smallest group which fails to satisfy the converse of Lagrange's theorem), it is easy to find this example.
• $C_5$ means?@the_fox
• Cyclic group of order $5$. Jan 11, 2019 at 5:46
• Do you know the fact that if $G$ has order $60$ and is not simple then it must have a normal subgroup of order $5$? Jan 11, 2019 at 6:07