# Showing H and Ha have same cardinality

• Mar 3rd 2010, 05:55 PM
ChrisBickle
Showing H and Ha have same cardinality
OK this was a question from a test ill post the question and then how i answered and i was hoping to get a "correct answer" and whether what i said was even close.

Let H be a subgroup of G. Show directly that H and Ha have the same cardinality.

I couldnt figure anything out proof wise so heres what i said:

H is a subgroup of G so H is finite and has elements h1,h2,...,hn all of which are distinct. Ha is the set {x in G | x = ha h in H} then Ha takes elements in H and multiplies (perform operation) on the right by a, since all elements in h are distinct then all h1a,h2a,...hna will be distinct so they must be the same size

I couldnt figure out a proof so i was trying to reason through it
• Mar 3rd 2010, 06:17 PM
tonio
Quote:

Originally Posted by ChrisBickle
OK this was a question from a test ill post the question and then how i answered and i was hoping to get a "correct answer" and whether what i said was even close.

Let H be a subgroup of G. Show directly that H and Ha have the same cardinality.

I couldnt figure anything out proof wise so heres what i said:

H is a subgroup of G so H is finite and has elements h1,h2,...,hn all of which are distinct. Ha is the set {x in G | x = ha h in H} then Ha takes elements in H and multiplies (perform operation) on the right by a, since all elements in h are distinct then all h1a,h2a,...hna will be distinct so they must be the same size

I couldnt figure out a proof so i was trying to reason through it

The map $h\mapsto ha$ is an obvious bijection between $H\,\,\,and\,\,\,Ha$

Tonio