Show isomorphism for element g in group G

In summary, the problem asks you to show that the map ig, defined as ig(x) = gxg' for x in G, is an isomorphism of G with itself. This means that you must show that ig is a bijection and that for all x, y in G, ig(xy) = ig(x)ig(y). To prove the latter, you must use the fact that g' is the inverse of g. To prove the bijection, you must show that ig is both injective and surjective, which can be done by using properties of inverse elements and cancellation.
  • #1
bennyska
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Homework Statement


let G be a group and let g be one fixed element of G. Show that the map ig, such that ig(x) = gxg' for x in G, is an isomorphism of G with itself.

Homework Equations


The Attempt at a Solution


not even really understanding the question. can someone break it down for me, and explain what the question is asking? am i trying to find a function, or rather show that i(x) preserves the structure?
 
Last edited:
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  • #2


Does it specify what is F? Or is g an element of G?

Also, does it specify what exactly is g' in ig(x) = gxg'?
 
  • #3


i'm sorry, it's supposed to be:
let G be a group and let g be one fixed element of G. Show that the map ig, such that ig(x) = gxg' for x in G, is an isomorphism of G with itself.
g' is the inverse of g, i think.
 
  • #4


I am assuming that g is an element of G and that g' is the inverse of g.

The problem is giving you a function and asking you to show that it is an isomorphism. To show that it is an isomorphism, you must show it is a bijection and that for all x, y in G, ig(xy) = ig(x)ig(y)
 
  • #5


so for the latter part:
ig(xy)=gxyg'
=gxg'gyg'
=ig(x)*ig(y)
is that right?
 
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  • #6


and for the bijection:

injection
let ig(x) = ig(y) for x,y in G. then
gxg'=gyg'
g'gxg'=g'gy'g inverses
xg'=yg'
x=y by right cancellation

surjection
let x,y in G. then
y=gxg'
yg=gxg'g
yg=gx
g'yg=g'gx
g'yg=x

is this right? i kind of feel like I'm using normal multiplication to do this, ignoring commutativity.
 

Related to Show isomorphism for element g in group G

What does it mean to show isomorphism for an element in a group?

Showing isomorphism for an element g in a group G means demonstrating that there exists a one-to-one and onto function between the set of elements in G and the set of elements in a different group H, preserving the group operation. This shows that the two groups are structurally equivalent.

Why is it important to show isomorphism for an element in a group?

Showing isomorphism for an element in a group is important because it allows us to understand the relationship between different groups and their structures. It also helps us identify patterns and similarities between groups, making it easier to study and solve problems related to them.

What are the steps to show isomorphism for an element in a group?

The steps to show isomorphism for an element g in a group G are as follows: 1. Define a function between G and another group H. 2. Show that the function is one-to-one and onto. 3. Prove that the function preserves the group operation. 4. Conclude that there exists an isomorphism between G and H.

How do you prove that a function preserves the group operation?

To prove that a function between two groups preserves the group operation, you need to show that for any two elements a and b in the first group, the result of the group operation on these elements is equal to the result of the group operation on the image of a and b under the function. In other words, f(a * b) = f(a) * f(b). This ensures that the structure of the group is maintained under the function.

Can two groups have multiple isomorphisms between them?

Yes, two groups can have multiple isomorphisms between them. This is because there can be different functions that satisfy the criteria for isomorphism, and thus, there can be multiple ways to show that the groups are structurally equivalent. However, it is important to note that two groups can have at most one isomorphism if they have a finite number of elements.

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