Question

(i) Let $G$ be a group of order $2^{m} k$, where $k$ is odd. Prove that if $G$ contains an element of order $2^{m}$, then the set of all elements of odd order in $G$ is a (normal) subgroup of G. (Hint. Consider $G$ as permutations via Cayley's theorem, and show that it contains an odd permutation.) (ii) Show that a finite simple group of even order must have order divisible by $4 .$

          (i) Let $G$ be a group of order $2^{m} k$, where $k$ is odd. Prove that if $G$ contains an element of order $2^{m}$, then the set of all elements of odd order in $G$ is a (normal) subgroup of G. (Hint. Consider $G$ as permutations via Cayley's theorem, and show that it contains an odd permutation.)
(ii) Show that a finite simple group of even order must have order divisible by $4 .$

Added by Ralph C.

Instant Answer

Solved by Expert Sam Stansfield

05/05/2023

Step 1

We want to show that $H$ is a normal subgroup of $G$. First, we show that $H$ is a subgroup of $G$. Show more…

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(i) Let $G$ be a group of order $2^{m} k$, where $k$ is odd. Prove that if $G$ contains an element of order $2^{m}$, then the set of all elements of odd order in $G$ is a (normal) subgroup of G. (Hint. Consider $G$ as permutations via Cayley's theorem, and show that it contains an odd permutation.) (ii) Show that a finite simple group of even order must have order divisible by $4 .$