Groups with a cyclic Sylow p-subgroup (1)

Posted: January 30, 2011 in Elementary Algebra; Problems & Solutions, Groups and Fields
Tags: ,

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.

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