Group which doesn't have subgroup of degree d||G|.

Asked Dec 17 '16 at 11:53

Active Dec 17 '16 at 11:53

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Today my teacher gave me the task: found finite group, which doesn't have subgroup of degree $d$ where $d$| |G|.

I thought about some special groups and actually about$\mathbb{Z}_{8}$, if there exist subgroup H with 2 elements , then this group contain $e$ and $a^{-1}$, but $a^{-1} = \overline{7-a} $ so there is 3 elements. Does it true that $\mathbb{Z}_{8}$ is good for this?

asked Dec 17 '16 at 11:53

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No, you need $d$ to not be a prime and you need the group to not be abelian, otherwise you cannot find an example. – Tobias Kildetoft Dec 17 '16 at 11:56
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The first counter example to Lagrange's theorem actually occurs in $A_4$. You may wish to check to see which divisor does not have a corresponding subgroup. – Evgeny T Dec 17 '16 at 11:57
You won't find examples among the abelian groups. – lhf Dec 17 '16 at 11:57
Possible duplicate of $A_4$ has no subgroup of order $6$? – Watson Dec 17 '16 at 12:28
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See also this, which explains why your example can't work. – Watson Dec 17 '16 at 12:29
@Watson it is NOT a duplicate of that question, please remove your comment. – Joffan Dec 17 '16 at 13:18

Group which doesn't have subgroup of degree d||G|.

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