# Group Theory Sylow Subgroup [closed]

What's an example of a group $$G$$ and an integer $$n$$ dividing $$|G|$$ with $$0 < n < |G|$$ such that $$G$$ has no subgroup of order $$n$$.

## closed as off-topic by ml0105, Randall, Saad, Dando18, Lord Shark the UnknownDec 10 '18 at 3:55

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• Please also consider adding what you have tried in solving the problem. In MSE bare problem statement questions (without attempt from the asker) are generally frowned upon. So please consider adding them. – user 170039 Dec 10 '18 at 3:50
• I will. Thanks for letting me know what MSE is like. – user624358 Dec 10 '18 at 6:31

In $$A_4$$,with $$|A_4|=12$$ are there subgroup of order $$6$$?
Suppose $$|G|$$ is simple non-cyclic, then $$G$$ is non-abelian (Use Cauchy's Theorem), then $$G$$ has even order (Apply Feit–Thompson theorem), hence $$G$$ has no subgroup of order $$|G|/2$$. So if $$G$$ is simple non-cyclic, then $$|G|=2n$$ and $$G$$ has no subgroup of order $$n.$$ In particular the monster group has no subgroup of order $$404008712397256437943229952480855378502877184000000000.$$
• You're right. My first inclination was just to mention $A_5,$ but then I thought why not apply it to a much larger class of groups? Even if it's totally unnecessary. – Melody Dec 10 '18 at 0:44