Is there any finite group G such that G has no subgroup of order $d_1$, $d_2$ but have a subgroup of order $d_1 d_2$

Question

Is there any finite group G with $d_1$, $d_2$ and $d_1\times d_2$ are proper dividors of $|G|$ such that G has no subgroup of order $d_1$, $d_2$ but have a subgroup of order $d_1\times d_2$

$S_5$ has subgroup $A_5$ of order 60 but does not have subgroup for order 15 and 30. But I am unable to find example for above case.

score 1 · Accepted Answer · answered Mar 08 '17 at 18:14

1

With $q=2^6$, let $h=\text{SL}(q)$ the group of $2\times 2$ matrices over $\mathbb{F}_q$ of determinant $1$. Its order is $64\cdot 63\cdot 65$. Take $d_1=8\cdot 7 \cdot 5$ and $d_2=8\cdot 9 \cdot 13$. For $G$ we can take the extension of $H$ by the automorphism induced by the field automorphism of order $2$.

The necessary checks are left to the reader.

answered Mar 08 '17 at 18:14

ancient mathematician

14,102
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Is this the smallest possible group? – SSSM Mar 08 '17 at 18:35
I don't think so, $\text{SL}_2(16)$ suitably extended will do. – ancient mathematician Mar 08 '17 at 18:49

Is there any finite group G such that G has no subgroup of order $d_1$, $d_2$ but have a subgroup of order $d_1 d_2$

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