Homomorphism of Groups

[Mathyes]

Would someone please help me verify my answer to this problem?

Problem: Consider the additive group \(\displaystyle {\Re ^2}\) and \(\displaystyle \Re \), where the operations are the usual ones. Let \(\displaystyle \pi :{\Re ^2} \to \Re \) be defined by \(\displaystyle \pi (x,y) = x + y\). Show that \(\displaystyle \pi \) is a homomorphism of groups.

Pf.

Let \(\displaystyle ({x_1},{y_1}),({x_2},{y_2}) \in {\Re ^2}\)

Show: \(\displaystyle \pi [({x_1},{y_1}) + ({x_2},{y_2})] = \pi ({x_1},{y_1}) + \pi ({x_2},{y_2})\)

\(\displaystyle \pi [({x_1},{y_1}) + ({x_2},{y_2})] = \pi [({x_1} + {y_1}) + ({x_2} + {y_2})]\) = \(\displaystyle ({x_1} + {y_1}) + ({x_2} + {y_2})\) = \(\displaystyle \pi ({x_1},{y_1}) + \pi ({x_2},{y_2})\) as desired

Therefore, \(\displaystyle \pi \) is a homomorphism of groups.

 

[pka]

Would someone please help me verify my answer to this problem?

Problem: Consider the additive group \(\displaystyle {\Re ^2}\) and \(\displaystyle \Re \), where the operations are the usual ones. Let \(\displaystyle \pi :{\Re ^2} \to \Re \) be defined by \(\displaystyle \pi (x,y) = x + y\). Show that \(\displaystyle \pi \) is a homomorphism of groups.

Pf.

Let \(\displaystyle ({x_1},{y_1}),({x_2},{y_2}) \in {\Re ^2}\)

Show: \(\displaystyle \pi [({x_1},{y_1}) + ({x_2},{y_2})] = \pi ({x_1},{y_1}) + \pi ({x_2},{y_2})\)

\(\displaystyle \pi [({x_1},{y_1}) + ({x_2},{y_2})] = \pi [({x_1} + {y_1}) + ({x_2} + {y_2})]\) = \(\displaystyle ({x_1} + {y_1}) + ({x_2} + {y_2})\) = \(\displaystyle \pi ({x_1},{y_1}) + \pi ({x_2},{y_2})\) as desired

Therefore, \(\displaystyle \pi \) is a homomorphism of groups.

What does one have to do in order to show a mapping is a group homomorphism? Have you done it?

 

[Mathyes]

I just did it and posted here!

But to answer your question,

I follow the definition of group homomorphism, which said, "A map \(\displaystyle \varphi \) of a group G into group G' is a homomorphism if the homomorphism property \(\displaystyle \varphi (ab) = \varphi (a)\varphi (b)\) holds for all \(\displaystyle a,b \in G\)."

Since the problem I was given above, asking me to consider the additive groups \(\displaystyle {\Re ^2}\) and \(\displaystyle \Re \) where \(\displaystyle \pi :{\Re ^2} \to \Re \) is defined as \(\displaystyle \pi (x,y) = x + y\).

So since \(\displaystyle {\Re ^2}\) map into \(\displaystyle \Re \), I choose \(\displaystyle ({x_1},{y_1}),({x_2},{y_2}) \in {\Re ^2}\). Then I set up my proof by letting \(\displaystyle \pi [({x_1},{y_1}) + ({x_2},{y_2})]\) (because of additive), and from there I must show that it is equal to \(\displaystyle \pi ({x_1},{y_1}) + \pi ({x_2},{y_2})\). Which is in a similar form of the definition of homomorphism \(\displaystyle \varphi (ab) = \varphi (a)\varphi (b)\) except I am asked to consider the additive groups which is in addition.

What I am not sure about is this step \(\displaystyle \pi [({x_1} + {y_1}) + ({x_2} + {y_2})]\) = \(\displaystyle ({x_1} + {y_1}) + ({x_2} + {y_2})\), which I based on the definition.

Thanks for your response,

 

[Steven G]

I just did it and posted here!

But to answer your question,

I follow the definition of group homomorphism, which said, "A map \(\displaystyle \varphi \) of a group G into group G' is a homomorphism if the homomorphism property \(\displaystyle \varphi (ab) = \varphi (a)\varphi (b)\) holds for all \(\displaystyle a,b \in G\)."

Since the problem I was given above, asking me to consider the additive groups \(\displaystyle {\Re ^2}\) and \(\displaystyle \Re \) where \(\displaystyle \pi :{\Re ^2} \to \Re \) is defined as \(\displaystyle \pi (x,y) = x + y\).

So since \(\displaystyle {\Re ^2}\) map into \(\displaystyle \Re \), I choose \(\displaystyle ({x_1},{y_1}),({x_2},{y_2}) \in {\Re ^2}\). Then I set up my proof by letting \(\displaystyle \pi [({x_1},{y_1}) + ({x_2},{y_2})]\) (because of additive), and from there I must show that it is equal to \(\displaystyle \pi ({x_1},{y_1}) + \pi ({x_2},{y_2})\). Which is in a similar form of the definition of homomorphism \(\displaystyle \varphi (ab) = \varphi (a)\varphi (b)\) except I am asked to consider the additive groups which is in addition.

What I am not sure about is this step \(\displaystyle \pi [({x_1} + {y_1}) + ({x_2} + {y_2})]\) = \(\displaystyle ({x_1} + {y_1}) + ({x_2} + {y_2})\), which I based on the definition.

Thanks for your response,

You compute pi of something in R^2 !!!
But you are computing \(\displaystyle \pi [({x_1} + {y_1}) + ({x_2} + {y_2})]\). Is \(\displaystyle \ [({x_1} + {y_1}) + ({x_2} + {y_2})]\) in R^2,

Being stuck is OK, but you are making a big mistake here.

 

[Steven G]

Would someone please help me verify my answer to this problem?

Problem: Consider the additive group \(\displaystyle {\Re ^2}\) and \(\displaystyle \Re \), where the operations are the usual ones. Let \(\displaystyle \pi :{\Re ^2} \to \Re \) be defined by \(\displaystyle \pi (x,y) = x + y\). Show that \(\displaystyle \pi \) is a homomorphism of groups.

Pf.

Let \(\displaystyle ({x_1},{y_1}),({x_2},{y_2}) \in {\Re ^2}\)

Show: \(\displaystyle \pi [({x_1},{y_1}) + ({x_2},{y_2})] = \pi ({x_1},{y_1}) + \pi ({x_2},{y_2})\)

\(\displaystyle \pi [({x_1},{y_1}) + ({x_2},{y_2})] = \pi [({x_1} + {y_1}) + ({x_2} + {y_2})]\) = \(\displaystyle ({x_1} + {y_1}) + ({x_2} + {y_2})\) = \(\displaystyle \pi ({x_1},{y_1}) + \pi ({x_2},{y_2})\) as desired

Therefore, \(\displaystyle \pi \) is a homomorphism of groups.

\(\displaystyle \pi [({x_1},{y_1}) + ({x_2},{y_2})]\)----can you re-write this to get it in the form of pi(something in R^2)????

 

[Mathyes]

Thank you for pointing it out.
Yes, big mistake.
It should be in the form of \(\displaystyle {\Re ^2} = \{ (x,y)|x,y \in \Re \} \).
I have to re-do my answer.
I guess I am having difficulty with the set up of the solution to get it in the form of \(\displaystyle \varphi (ab) = \varphi (a)\varphi (b)\), but for this specific problem. I know that I need to set it up correctly from the left side of the equal sign and then show that it leads to give me the right side.

Here is what I am thinking:

\(\displaystyle \pi [({x_1},{y_1}),({x_2},{y_2})]\) or \(\displaystyle \pi (x,y)\). Any way will continue to work on it.

 

[Steven G]

Thank you for pointing it out.
Yes, big mistake.
It should be in the form of \(\displaystyle {\Re ^2} = \{ (x,y)|x,y \in \Re \} \).
I have to re-do my answer.
I guess I am having difficulty with the set up of the solution to get it in the form of \(\displaystyle \varphi (ab) = \varphi (a)\varphi (b)\), but for this specific problem. I know that I need to set it up correctly from the left side of the equal sign and then show that it leads to give me the right side.

Here is what I am thinking:

\(\displaystyle \pi [({x_1},{y_1}),({x_2},{y_2})]\) or \(\displaystyle \pi (x,y)\). Any way will continue to work on it.

For some reason I am getting tired of seeing this problem. I hope no one minds me showing the key step this student is not seeing.

Just think of (x1,y1) + (x2, y2) as (x1+x2, y1+y2) and now it is in R^2. OK, now go and finish up this proof.
Also show us the completed proof.

 

[Mathyes]

Thank you for the key step,

Here is the proof:

Let \(\displaystyle ({x_1} + {x_2}),({y_1} + {y_2})\) in \(\displaystyle {\Re ^2}\). Then \(\displaystyle \pi [({x_1} + {x_2}),({y_1} + {y_2})] = ({x_1} + {x_2}) + ({y_1} + {y_2})\) = \(\displaystyle \pi ({x_1},{x_2}) + \pi ({y_1},{y_2})\) as desired.
Therefore, \(\displaystyle \pi \) is homomorphism of groups.

 

[pka]

Here is the proof:
Let \(\displaystyle ({x_1} + {x_2}),({y_1} + {y_2})\) in \(\displaystyle {\Re ^2}\). Then \(\displaystyle \pi [({x_1} + {x_2}),({y_1} + {y_2})] = ({x_1} + {x_2}) + ({y_1} + {y_2})\) = \(\displaystyle \pi ({x_1},{x_2}) + \pi ({y_1},{y_2})\) as desired.
Therefore, \(\displaystyle \pi \) is homomorphism of groups.

I too am tried of this, so here is my take on it,
Elements of \(\displaystyle \Re^2\) look like \(\displaystyle (x,y)\) and elements of \(\displaystyle \Re^1\) look like \(\displaystyle x\)
Addition in the group \(\displaystyle (a,b)+(x,y)=(a+x,b+y)\)

If \(\displaystyle \Pi:\Re^2\to\Re^1\) as \(\displaystyle \Pi: (x,y)\mapsto x+y\) then
\(\displaystyle \begin{align*}\Pi((a,b)+(c,d))&=\Pi(a+c,b+d)\\&=(a+c)+(b+d)\\&=(a+b)+(c+d)\\&=\Pi(a,b)+\Pi(c,d) \end{align*}\)

Please study the placement of commas. Than tells us about which group the term is in.