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Prove the following is a homomorphism and describe its kernel.

The function $f: \mathbb{RxR} \rightarrow \mathbb{R} $ given by $f(x,y)=x+y$

I just want someone to confirm my answer:

My answer:

Since $\mathbb{RxR}$ and $\mathbb{R}$ are both groups and there exists a function $f$, then for any two elements $x$ and $y$:

$$f(x,y)=x+y=f(x)+f(y)$$

As for kernel, would it be $K=\{(-x,-y) \in G | f(-x,-y)=0\}$

kero
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    this can't possibly be correct. How do you go from $x+y$ to $f(x)+f(y)$? what is $f(x)$ or $f(y)$ for that matter???? (answer: nothing, as it is undefined). Look closely at the domain for $f$. What is the group operation on the domain? – Ittay Weiss Apr 05 '15 at 04:34

2 Answers2

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(assuming additive notation throughout)

To show that $f$ is a homomorphism, you need to show $f(z_1 + z_2) = f(z_1) + f(z_2)$ for all $z_1$ and $z_2$ in the group $\mathbb{R} \times \mathbb{R}$. Since the structure on this group is coordinatewise addition, the statement to be proved is of the form $f(x_1+x_2, y_1+y_2) = f(x_1,y_1)+f(x_2,y_2)$.

For the kernel, you should aim for a more intuitive description. The definition is 'those $(x,y)$ such that $f(x,y) = 0$' The definition of $f$ should make it clear that this can be further simplified.

Rolf Hoyer
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  • Alright! Let's see if I get it this time: $f(x,y)=f(x_1,y_1)+f(x_2,y_2)=f(x_1+x_2)+f(y_1+y_2)=(x_1+x_2)+(y_1+y_2)=f(x)+f(y)$ – kero Apr 05 '15 at 05:01
  • Would I have to state what $x_1$ and $x_2$ or is that understandable? Also, i'm not quite understanding what you are trying to say for the kernel part. Do you mean that the kernel is (0,0)? – kero Apr 05 '15 at 05:02
  • I am having trouble parsing your argument. Since $f$ is a binary function, I don't know what $f(x_1+x_2)$ or $f(x)$ could be, for instance. Are $x$ or $y$ ordered pairs or single real numbers? – Rolf Hoyer Apr 05 '15 at 05:05
  • I guess I'm being being deliberately cryptic about the kernel part, you shouldn't take my words as indicating any particular answer. – Rolf Hoyer Apr 05 '15 at 05:07
  • f(x,y)=f(x1,y1)+f(x2,y2)=f(x1+x2)+f(y1+y2)=(x1+x2)+(y1+y2)=f(x)+f(y‌​) – Justin Apr 05 '15 at 05:11
  • @RolfHoyer, the OP is confusing binary with unary functions. Writing $f(z_1)$ is only going to add to this problem. – vadim123 Apr 05 '15 at 05:13
  • I'm confused. Why are you taking $n$ pairs now? What does the pair $(x,y)$ have to do with the pairs $(x_i, y_i)$? The right hand side of your argument is plugging a single real number into $f$, which is invalid. – Rolf Hoyer Apr 05 '15 at 05:15
  • OK, maybe things are more clear now? – Rolf Hoyer Apr 05 '15 at 05:18
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Homomorphism: $$f((x,y)+(x',y'))=f((x+x',y+y'))=x+x'+y+y'=(x+y)+(x'+y')=\\=f((x,y))+f((x',y'))$$

Kernel: $$0=f((x,y))=x+y$$ This holds exactly when $y=-x$, i.e. $Ker(f)=\{(x,-x):x\in \mathbb{R}\}$

vadim123
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