Need help understanding the concept Let G1 = , G2 = 0, x(multiplication) f:G1-- G2 defined by f(x) = e^x Are G1, G2 isomorphic? the answer is yes and the logic is $e^{x+y}=e^x *e^y$. This preserves groups structure, preserves multiplication table. Please, explain me what it mean.

Question

Need help understanding the concept Let G1 = , G2 = 0, x(multiplication)> f:G1--> G2 defined by f(x) = e^x Are G1, G2 isomorphic? the answer is yes and the logic is $e^{x+y}=e^x *e^y$. This preserves groups structure, preserves multiplication table. Please, explain me what it mean.

anonymous · Answer

$G_1=\langle\mathbb{R},+\rangle$ and $G_2=\langle \mathbb{R}_{+},\times\rangle$ ? where $\mathbb{R}_+$ is positive reals only?

anonymous · Answer

Yes, Sir.

anonymous · Answer

Okay, in this case, an isomorphism is a bijective function $f:G_1\to G_2$ such that $f(u+v)=f(u)\times f(g)$ for all $u,v\in G_1$, which is the case for the exponential function.

anonymous · Answer

$$f(x+y)=e^{x+y}$$
$$f(x)\times f(y)=e^x\times e^y$$

anonymous · Answer

Yes, I got it so far, but "it preserves groups structure" mean??

anonymous · Answer

* the $f(g)$ should say $f(v)$, sorry.

anonymous · Answer

Well, what is a group? It's a set with an operation that is closed under the operation, associative, has an identity element, and is invertible.

$f$ is closed under the operation, since $e^x$ and $e^y$ are real, and so is the product $e^{x+y}$.

$f$ is associative, since $e^{(x+y)+z}=e^{x+(y+z)}$ and $(e^xe^y)e^z=e^x(e^ye^z)$.

The identity element is 0, since $e^{0+x}=e^0e^x=e^x$.

I'm not sure about invertibility, but I think it has to do with the logarithm function.

anonymous · Answer

You mean after working on the function f, the range of f = set of elements in G2, right?

anonymous · Answer

Actually, that identity element might not be right... which means the inverse element has less to do with logs.

Yes, the range is the set associate with $G_2$.

anonymous · Answer

I think I got it. Thanks a lot, Professor. :)

anonymous · Answer

I watch this video, this problem is at 21:23 https://www.youtube.com/watch?v=mwcNETa0KFI

anonymous · Answer

Maybe the identity is just 1, which makes the inverse element $\dfrac{1}{x}$ for $x\in G_1$, $x\not=0$. You don't have to worry about that for $x\in G_2$ though.

anonymous · Answer

And you're welcome, but I'm no professor :P

anonymous · Answer

Ok, I remember that once you said "I..... , I am bad professor" . That's why I think you are. :)

anonymous · Answer

Hmm, I don't remember saying anything like that... Oh well, happy to help :)

anonymous · Answer

:)