Processing math: 100%
Jump to content

Recommended Posts

Posted
  Quote
Let G be any group, and C'=\{a \in G: (ax)^{2}=(xa)^{2} for every x \in G\}.

Prove that C' is a subgroup of G.

This exercise is from chapter 5 of A Book Of Abstract Algebra.

I've been studying this book by reading, so this is not homework. Please help. Here is my attempt:

If C' is a subgroup,

(abx)^{2}=(xab)^{2}

and

(a^{-1}x)^{2}=(xa^{-1})^{2}

for every a and b in C'.

It is obvious that Z(G) is included in C'. If we have a group K=\{a,b \in K: a=a^{-1}, ab \neq ba \},

then Z(K)=\emptyset even though (ab)^{2}=(ba)^{2} for every element in K. Thus C' is not necessarily Abelian.

Posted

Solved

(a(bx))^{2}=((bx)a)^{2}=(b(xa))^{2}=((xa)b)^{2}=(xab)^{2}

and

(a^{-1}x)^{2}=(a^{-1}xa^{-1}a)=((a^{-1}xa^{-1})a)^{2}=(a(a^{-1}xa^{-1}))^{2}=(xa^{-1})^{2}.

Thus C' is closed with respect to multiplication and inverses, and is a subgroup of G.

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.