The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

Subgroups prove that question FP3

Let G be a group, and let g be a fixed element of G. Prove that the set of elements which commute with g that is H={x belongs to G : gx=xg} is a subgroup of G.

I dont understand why it would be a subgroup because xg or gx might not equal g or x, so its not a closed group?
Reply 1
Avatar for davros
davros
16
Original post by physics4ever
Let G be a group, and let g be a fixed element of G. Prove that the set of elements which commute with g that is H={x belongs to G : gx=xg} is a subgroup of G.

I dont understand why it would be a subgroup because xg or gx might not equal g or x, so its not a closed group?


You seem to be a bit confused about what you're trying to prove!

Closure means that if x and y are both in H then xy and yx are both in H too (it might be that they're the same element, but you can't assume that to start with!).
Reply 2
Avatar for und
und
16
Original post by physics4ever
Let G be a group, and let g be a fixed element of G. Prove that the set of elements which commute with g that is H={x belongs to G : gx=xg} is a subgroup of G.

I dont understand why it would be a subgroup because xg or gx might not equal g or x, so its not a closed group?

Suppose we have x and y in H. Then all we can say for certain is that x and y both commute with g. But it is easy to show that xy and yx also commute with g, so the elements xy and yx are also in H by definition.

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you