Abstract Algebra #1

Algebra Level 3

Every proper subgroup of a nonabelian group is nonabelian.

False True

Mike Argus by Mike Argus , 21, USA

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.

3 solutions

Mike Argus
Sep 2, 2015

We will provide a counterexample to show the statement is false.

Let G be a nonabelian group with identity element e. Consider the trivial subgroup {e}, which is clearly abelian. As G is nonabelian it follows that G cannot be the trivial subgroup {e}, and so, {e} is in fact a proper abelian subgroup of G.

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Ravi Dwivedi
Sep 4, 2015

We note the following counterexample to prove that the statement is false

Let G G be any non abelian group .

consider an element a ∈ G a \in G

Then the subgroup of G G generated by this element a a is

< a > = a n : n ∈ Z <a> = {a^n: n \in Z} which is abelian.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Of course, we should also check that < a > ≠ G <a> \neq G , but that's obvious.

Curtis Clement
Sep 2, 2015

I think I'm right in saying that the inverse on it's own is sufficient to form a subgroup. If not then in some groups you may have the inverse (say I \ I ) and an element a such that a = a − 1 ⟹ a a = I \ a = a^{-1} \implies\ aa = I

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Nope. For example, in the integers with addition, { 1 , − 1 } \{ 1, -1 \} is a set in which every element has it's inverse, but it's not a subgroup since it doesn't contain the identity 0.

Calvin Lin Staff - 5 years, 9 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...