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It is requested to proof exercise $3.30$, $(ii)$ from J. Rotman, An introduction to the theory of groups, which states that a finite simple group $G$ of even order greater than $4$ must have order divisible by $4$.

Number $(i)$ states that: for a group $G$ of order $2^m k$ where $k$ is an odd number. If $G$ has exactly one element of order $2^m$, then the set of all elements with odd order in $G$ is a normal subgroup of G.

Here is what has been done:

Since $|G|$ is even, $2$ divides $|G|$. As $G$ is finite group, it exists $m \in \mathbb{Z}$ such that $2^m$ divides $|G|$ but $2^{m+1}$ do not divides $|G|$. Thus, $|G| = 2^m k$ for some odd number $k$.

If $m > 1$ or $|G| = 4$, it's done. So, let's suppose that $m = 1$ and $|G| > 4$. Then, $|G| = 2 k$. It is known that for a group $G$ of even order, there is $a \in G$ such that $a^2 = 1$. Therefore, the set $N$ of all elements with odd order is normal subgroup of $G$.

Since $G$ is simple, $N = \{1\}$ or $N = G$. As $a$ has even order, it is not the case that $N = G$. Thus, $N = \{1\}$ which implies that the neuter element is the only element of odd order.

Since $|G| > 4$, $G \setminus \{1, a\} \ge 3$. Let $h_1, h_2 \in G$ such they have even order.

Case $1)$. If both of them have order $2$, then $H = \{1, a, h_1, h_2\} \le G$ and $4$ divides de order of $G$.

Case $2)$. WLOG, suppose that $h_1 = h_1^{-1}$ and $h_2 \neq h_2^{-1}$. Since $\langle h_2 \rangle$ is cyclic of even order, it exists $k_1 \in \langle h_2 \rangle$ such that $k_1^2 = 1$. Therefore, $H = \{1, a, h_1, k_1\} \le G$ and $4$ divides de order of $G$.

Case $3)$. Suppose that $h_1 \neq h_1^{-1}$ and $h_2 \neq h_2^{-1}$. Then, it exists $k_1 \in \langle h_1 \rangle$ and $k_2 \in \langle h_2 \rangle$ such that $k_1^2 = k_2^2 = 1$. Therefore, $H = \{1, a, k_1, k_2\} \le G$ and $4$ divides de order of $G$.

How can it be guaranteed that $|H| = 4$? That is, in case $2)$ that $a \neq k_1$ and $h_1 \neq k_1$ and in case $3)$ that $a \neq k_1$, $a \neq k_2$ and $k_1 \neq k_2$?

EDIT 1:

I found the problem: since every element of $G\setminus \{1\}$ is even, $G$ is a $2-group$, which implies that the order of $G$ is a power of $2$. The contradiction comes by supposing that $m = 1$.

Blue Tomato
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  • Note: proof =/= prove – Shaun May 01 '23 at 22:10
  • I found a valuable article: https://math.stackexchange.com/questions/2681388/elements-of-even-order-in-a-finite-group – Blue Tomato May 01 '23 at 22:27
  • That's not an article – Shaun May 01 '23 at 23:43
  • @Shaun it has the idea, that's the point Mr. Shaun. – Blue Tomato May 01 '23 at 23:49
  • I know, I know; just calling it an article is a little far. – Shaun May 02 '23 at 00:12
  • In your Case.(1) why is $H$ closed? – MathFail May 02 '23 at 00:19
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    A much shorter argument: If $G$ has order $2k$ with $k>1$ odd, then we have a homomorphism $G\to S_{2k}$, given by the action of $G$ on itself, and the sgn homomorphism $S_{2k}\to {\pm1}$. The composition $G\to {\pm1}$ is surjective, as any element of even order acts on $G$ as an odd number of cycles of even length. Thus as $k>1$, we conclude $G$ cannot be simple. This argument has appeared before: e.g. here: https://math.stackexchange.com/questions/488053/on-simple-groups-g-where-2-mid-g-4-not-mid-g – tkf May 02 '23 at 05:22
  • @MathFail we cannot guaranteed that sir. Thanks – Blue Tomato May 02 '23 at 13:18

1 Answers1

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First of all, $G$ is not a $2$-group. Now on the contrary, assume that $G$ is a simple of order $2m$, where $m>1$ is an odd integer. There are many ways to show that for any group of order $2n$, where $n>1$ is odd, there exists a normal(unique) Hall $2'$-subgroup(one of them is mentioned in the comments), a contradiction.

Another way to show the existence of such Hall subgroup is using the following lemma:
If Sylow $p$-subgroup of $G$ is cyclic, where $p$ is the smallest prime divisor of $|G|$, then $G$ has a normal $p$-complement. Here obviously Sylow $2$-subgroup of $G$ is cyclic.

We actually have the much stronger statement:
For any non-abelian simple group $G$, either $8\mid |G|$ or $12\mid |G|$.

Ash
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