I want to prove the following:

Let $G$ be a group of order $2^nm$, where $m$ is odd, having a cyclic Sylow $2$-subgroup. Then $G$ has a normal subgroup of order $m$.


We will show that $G$ has a subgroup of order $m$. Let $\theta:G\to \text{Sym}(G)$ be the homomorphism which is defined as $$ \theta(g)=t_g, \quad \forall g\in G $$ where $t_g:G\to G$ is defined as: $$ t_g(x)=gx, \quad \forall x\in G. $$ Let $g$ be an element of order $2^n$ in $G$. (There exists such an element since $G$ has a cyclic Sylow $2$-subgroup.)

The cyclce representation of $t_g$ is a product of $m$ disjoint cycles each of length $2^n$. Therefore $t_g$ is an odd permutation.

Thus the homomorphism $\epsilon\circ \theta:G\to\{\pm 1\}$, where $\epsilon:\text{Sym}\to\{\pm 1\}$ is the sign homomorphism, is a surjection.

By the First Isomprphism Theorem, we conclude that the kernel $K$ of $\epsilon\circ \theta$ is of order $2^{n-1}m$.

If $n$ were equal to $1$ then we are done.

If $n>1$ then note that any Sylow $2$-subgroup of $K$ is also cyclic. This is because each Sylow $2$-subgroup of $K$ is contained in a Sylow $2$-subgroup of $G$, where the latter is cyclic.

Now we can inductively show that $K$ has a subgroup of order $m$.

What I am struggling with is showing the normality.

Can anybody please help me with this.



Claim: If a group of order $2^nm$, $m$ odd, has cyclic Sylow-2 subgroup, then $G$ has unique subgroup of order $m$.

Proof: Induction on $n$- for $n=1$, as you showed, the kernel $K$ has order $m$, which is a normal subgroup. If there is another subgroup $H$ of order $m$, then the product $KH$ is a subgroup (since $K\trianglelefteq G$) of odd order (equal to $|H|.|K|/|H\cap K|$), and is bigger than $m$ (since $H\neq K$), which is impossible since the largest odd order dividing $|G|$ is $m$.

Suppose theorem is proved for groups of order $2m, 2^2m, \cdots, 2^{n-1}m$ (containing cyclic Sylow-$2$). Let $|G|=2^nm$, with cyclic Sylow-2. As you noted, kernel $K$ has order $2^{n-1}m$, which contains unique subgroup of order $m$ (by induction), say it is $L$. Thus $L$ is characteristic in $K$ and as $K$ is normal in $G$, it follows that $L\trianglelefteq G$.

Again, as in previous paragraph (at the starting of proof), we can conclude that $G$ has unique subgroup of order $m$, a stronger conclusion than you expected.

(Very Simple Exercise: $H$ is characteristic in $K$ and $K\trianglelefteq G$ $\Rightarrow$ $H\trianglelefteq G$. Just apply definition otherwise see this)

  • 1
    $\begingroup$ Superb answer! Thanks! $\endgroup$ – caffeinemachine Sep 4 '14 at 9:33
  • $\begingroup$ As a critique, in your inductive step, I don't believe you are proving enough. Not only do you have to show that $L$ is normal in $G$, you must show that $L$ is characteristic in $G$ for the induction follow. $\endgroup$ – Frank Wang Dec 2 '15 at 9:21

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