Groups without using Sylow

Question

I am going over older papers that dont have solutions and again wanted to ask for some help. Previously in this exercise I proved the Recognition Criterion and the fact that if every element of $G$ has order at most 2, then $G$ is abelian and is isomorphic to $C_2$x$C_2$...x$C_2$ (n lots of $C_2$ for some n). I think I must somehow use this but I dont quite see how

Let G be a finite group such that $\vert G\vert = 2n$. Assume $G$ has a subgroup $K \subset G$ such that $\vert K \vert = n$. WITHOUT USING SYLOWS THEOREMS show the following:

i) $\exists q \in G$ such that $ord(q)=2$

ii) $K$ is normal and $G \cong K \rtimes_hQ $ for some homomorphism $h: Q \to Aut(K)$

I am really in a bind here. The case when the $n$ in $2n$ is prime was covered in our lectures but when $n$ is not prime, the situation obviously changes.

Is $n$ odd? Otherwise ii) is not necessarily true. For example, a cyclic group of order $4$ is not isomorphic to a semidirect product. — Derek Holt, Mar 31 '18 at 14:18

Dietrich Burde · Accepted Answer · 2018-03-31T14:57:30.793

1

Question $(i)$ follows from Cauchy's Theorem for $p=2$. In question $(ii)$, if $K$ has index $2$, then it is normal:

Subgroup of index 2 is Normal

edited Mar 31 '18 at 14:57

answered Mar 31 '18 at 12:59

Dietrich Burde

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Groups without using Sylow

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