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Let $G$ be a finite group. Assume that every subgroup of order 4 in $G$ is cyclic (as happens if $G$ is a cyclic group or a generalized quaternion group). It seems to me that it should follow that a Sylow 2-subgroup of $G$ is either a cyclic group or a generalized quaternion group. However, I cannot find a proper reference for this, nor have I been successful on proving it.

I think that this may be well-known, does anyone know a reference I could have a look at?

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    $\begingroup$ Let $S$ be a Sylow $2$-subgroup of $G$. Then $Z(S)$ contains an element $t$ of order $2$. Furthermore, $t$ is the only involution (element of order two) of $S$, for if there were another involution $u \in S$, then $\langle t,u \rangle$ is a non-cyclic subgroup of $S$ of order $4$. It is indeed true that a $2$-group which contains a unique involution is either cyclic or generalized quaternion. This proved in many group theory texts, for example Finite Groups by D. Gorenstein. $\endgroup$
    – Geoff Robinson
    Commented Oct 12, 2020 at 20:04
  • $\begingroup$ Thank you @GeoffRobinson! Indeed the proof was very easy. $\endgroup$
    – quaternion
    Commented Oct 13, 2020 at 5:40

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