$G$ 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.

However, can you prove that this is the only exception?

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  • $\begingroup$ $C_5$ means?@the_fox $\endgroup$ – cmi Jan 11 '19 at 5:45
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    $\begingroup$ Cyclic group of order $5$. $\endgroup$ – the_fox Jan 11 '19 at 5:46
  • $\begingroup$ No I can not see why it is the only exception @the_fox $\endgroup$ – cmi Jan 11 '19 at 5:57
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    $\begingroup$ Do you know the fact that if $G$ has order $60$ and is not simple then it must have a normal subgroup of order $5$? $\endgroup$ – the_fox Jan 11 '19 at 6:07

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