Question

Is (R/{0}, *) isomorphic to (R, +)?

Answer 1

Of course not :)

Answer 2

But you're probably not gonna take my word for it, and you require an explanation, am I right? ^.^

Answer 3

>.>
Anyway...
Let's let G = <R\{0} , *> H=<R,+>
And suppose
\[\Large G\cong H\]

Answer 4

Well, since they're isomorphic, there's a function
\[\Large \phi:G\rightarrow H\]such that for any two elements g_1 and g_2 in G
\[\Large \phi(g_1 \times g_2)=\phi(g_1)+\phi(g_2)\]

Answer 5

Also, since it's an isomorphism, identity in G must be mapped to the identity in H.
\[\Large \phi(1)=0\]

Answer 6

So, let's get started :)
We don't know what \[\Large \phi(-1) \]is, but let's make it a real number x.
\[\Large \phi(-1)=x\]
Now, because of the isomorphism
\[\Large \phi(-1 \times -1)=\phi(-1)+\phi(-1)=x + x=2x\]

Answer 7

But
\[\Large -1 \times -1=1\]implying\[\Large 2x=\phi(-1\times-1)=\phi(1)=0\]Therefore \[\Large 2x=0\]\[\Large x=0\]

Answer 8

Then that means
\[\Large \phi(-1)=x=0\]
Which means
\[\Large \phi(-1)=0=\phi(1)\]
Means
\[\LARGE \phi:G\rightarrow H\]cannot be bijective, and thus, cannot be an isomorphism

Case closed ^.^