# Thread: Group Theory-Show no infinite soluble group has a composition series

1. ## Group Theory-Show no infinite soluble group has a composition series

Hi.....i'm stuck on the following question;

Show that no infinite soluble group has a composition series.

The hint for the question was to try by contradiction.

So i tried the following;

Suppose G is an infinite soluble group with a composition series.
G is soluble so all composition factors are cyclic and of prime order.

I think there will be a contradiction because G is infinite...but i just don't know

2. Originally Posted by Louise
Hi.....i'm stuck on the following question;

Show that no infinite soluble group has a composition series.

The hint for the question was to try by contradiction.

So i tried the following;

Suppose G is an infinite soluble group with a composition series.
G is soluble so all composition factors are cyclic and of prime order.

I think there will be a contradiction because G is infinite...but i just don't know

You almost have it: there's a (finite) series for G all of which factors are finite cyclic of prime order. But then, since a finite extension of a finite group is itself finite, we see G must be finite...(if $\displaystyle G >= G_1 >= G_2 >=....>=G_n= 1$ is the cyclic series, then $\displaystyle G_n$ is finite, so $\displaystyle G_{n-1}$ is finite since $\displaystyle G_n\,\, and\,\, G_{n-1}/G_n$ are finite, etc...)