Subgroup and quotient

Algebra Level 4

Let G G be a group, and N N a normal subgroup. Which of the following statements is/are always true?

I. If N N is finite and G / N G/N is finite, then G G is finite.
II. If N N is finite and cyclic and G / N G/N is finite and cyclic, then G G is finite and cyclic.
III. If N N is abelian and G / N G/N is abelian, then G G is abelian.


Notation:

  • A finite cyclic group is a group that is isomorphic to Z n , {\mathbb Z}_n, the integers mod n , n, for some n . n.
  • An abelian group is a group whose operation is commutative: x y = y x x * y = y * x for all x , y G . x,y \in G.
I only III only II and III II only I and II I and III

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Patrick Corn
Jun 25, 2016

I is true: the elements of G G are partitioned into disjoint cosets g N . gN. Since N N is finite, the cosets are finite. Since G / N G/N is finite, there are finitely many cosets. So G G is finite (and indeed G = N G / N |G| = |N| \cdot |G/N| ).

II and III are false. Here is a counterexample for both. Let G = S 3 , G = S_3, the symmetric group on three symbols, and N = A 3 = { i d , ( 123 ) , ( 132 ) } . N = A_3 = \{id,(123),(132)\}. Then N N is cyclic (isomorphic to Z 3 {\mathbb Z}_3 ) and G / N G/N is cyclic (isomorphic to Z 2 {\mathbb Z}_2 ) but G G is neither cyclic nor abelian.

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...