Groups with a cyclic Sylow p-subgroup (1)

Posted: January 30, 2011 in Elementary Algebra; Problems & Solutions, Groups and Fields
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Let $G$ be a finite group. Let $\text{Sym}(G)$ be the group of permutations on $G.$ Define the map $\varphi: G \longrightarrow \text{Sym}(G)$ by $\varphi(g)(x)=gx,$ for all $g, x \in G.$ It is easy to see that $\varphi$ is a group homomorphism:

$\varphi(g_1g_2)(x)=g_1g_2x=\varphi(g_1) \varphi(g_2)(x),$

for all $g_1,g_2, x \in G.$ Thus $\varphi(g_1g_2)=\varphi(g_1) \varphi(g_2),$ i.e. $\varphi$ is a group homomorphism.

Lemma 1. If the order of $a \in G$ is $d,$ then $\varphi(a)$ is a product of $|G|/d$ disjoint cycles of length $d.$

Proof. It is pretty obvious: choose $g \in G.$ Then, since $\varphi(a)(g)=ag,$ we will have

$\varphi(a)(ag)=a^2g, \ \varphi(a)(a^2g)=a^3g, \cdots , \ \varphi(a)(a^{d-1}g)=a^dg=g.$

So

$\sigma_g = (g \ \ ag \ \ \cdots \ \ a^{d-1}g)$

is a cycle of length $d$ in the decomposition of $\varphi(a).$ Now let $h \in G \setminus \{a,ag, \cdots , a^{d-1}g \}.$ Then $\sigma_{h}$ is another cycle of length $d$ in the decomposition of $\varphi(a).$ Clearly $\sigma_g$ and $\sigma_{h}$ are disjoint. We can continue this process until no element in $G$ is left. $\Box$

Problem 1. Let $G$ be a group with $|G|=2^nm,$ where $m$ is odd. Suppose that $G$ has a cyclic Sylow $2$-subgroup. Prove that $G$ has a (normal) subgroup $N$ of index $2.$

Solution. Define the map $\psi : G \longrightarrow \{1,-1\},$ by $\psi(g) = \text{sgn}(\varphi(g)),$ for all $g \in G.$ Here $\{1,-1\}$ is considered as the multiplicative group of order 2 and $\text{sgn}(\varphi(g))$ is the signature of the permutation of $\varphi(g).$ Clearly $\psi$ is a group homomorphism because it is the composite of two group homomorphisms. If $e$ is the identity element of $G,$ then obviously $\psi(e)=1.$ Now, if $P= \langle a \rangle$ is a cyclic subgroup of $G$ of order $2^n,$ then $a$ has order $2^n$ and thus, by Lemma 1, $\varphi(a)=\sigma_1 \sigma_2 \cdots \sigma_m,$ where each $\sigma_i$ is a cycle of length $2^n.$ Thus $\text{sgn}(\sigma_i)=-1$ and hence $\text{sgn}(\varphi(a))=(-1)^m=-1,$ because $m$ is an odd integer. So we have proved that $\psi$ is onto. Let $N = \ker \psi.$ Then $G/N \cong \{1,-1\}$ and therefore $[G:N]=|G/N|=2. \ \Box$

Lemma 2. Let $G$ be a finite group and let $N$ be a normal subgroup of $G.$ Let $p$ be a prime divisor of $|G|$ and suppose that $P$ is a Sylow $p$-subgroup of $G.$ Then $N \cap P$ is a Sylow $p$-subgroup of $N.$

Proof. Let $|P|=p^n$ and $|N|=p^rs,$ where $\gcd(p,s)=1.$ Since $N \cap P$ is a subgroup of both $P$ and $N,$ we have $|N \cap P|=p^t,$ for some $t \leq r.$ So in order to show that $N \cap P$ is a Sylow $p$-subgroup of $N,$ we only need to show that $t \geq r.$ Now,

$\displaystyle |PN|=\frac{|P| \cdot |N|}{|N \cap P|}=p^{n+r-t}s.$

But $PN$ is a subgroup of $G$ because $N$ is normal. Thus $n+r-t \leq n$ and so $t \geq r. \ \Box$

We will continue this discussion in the next post.