Question

If is a group such that every one of its Sylow subgroups (for every prime ) is cyclic and normal, prove that is a cyclic group.

Short answer

It is proved that is a cyclic group.

Step-by-step solution

Definition of Subgroup

Let (G,*) be a group under binary operation, say *. A non-empty subset H of G is said to be a subgroup of G, if (H,*) is itself a group.

In other words, if$a,b\in H$ and a,b$\in$H, then role="math" localid="1653284272887" $a{b}^{-1}\in H$.

Definition of Abelian Group

A group G is called abelian if ab=ba, $\forall$a,b $\in$G.

A group G is called cyclic if there exists an element a$\in$G, such that every element of G can be written as $a^n$, for some $n\in Z$.

Let G be a group, and every Sylow subgroup is cyclic and normal for every prime p.

Corollary 9.16 states that let G be a finite group and K a Sylow p-subgroup for some prime p. Then, K is normal in G if and only if K is the only Sylow p-subgroup in G.

It follows that for every prime divisor p of $\left|G\right|$, the group G has a unique Sylow p -subgroup $G_P$.

To prove G is a cyclic group

Now, by Exercise 9.3.13, all subgroups of a group are normal, we have $G\cong \prod _{p\left|G\right|}{G}_P$:

FromLemma 9.8, it follows that:

If $G_P$are cyclic, then G is also a cyclic group.

Hence, G is a cyclic group.