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$G$ 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?

the_fox
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cmi
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  • Maybe some similar questions will inspire you: https://math.stackexchange.com/q/94548/ https://math.stackexchange.com/q/2594060 https://math.stackexchange.com/q/2911121 – Trevor Gunn Jan 11 '19 at 05:18
  • Okk I am trying to get some idea from them@TrevorGunn – cmi Jan 11 '19 at 05:20
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    Use Sylow theorems. https://en.wikipedia.org/wiki/Sylow_theorems Factor 60 and look at the Sylow subgroups. – Pratyush Sarkar Jan 11 '19 at 05:23

1 Answers1

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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?

the_fox
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