Determining which maps are isomorphisms 1) Let $G=G'=\{(a,b)\mid a,b \in \mathbb{R}, a, b \ne 0\}$ with group operation $(a_1,b_1)(a_2,b_2)=(a_1a_2,b_1b_2)$. Let $\phi (a,b) = (b^{-1}, ab^2)$.
My solution: 
1-1: Suppose $(b_1^{-1},a_1b_1^2)=(b_2^{-1},a_2b_2^2)$. Then $b_1^{-1} = b_2^{-1}$ will imply $b_1 = b_2$ since they are non-zero reals and thus can be represented as equal fractions. Since  $b_1 = b_2$, $a_1b_1^2 = a_2b_2^2$ implies $a_1 = a_2$ since we can divide out the b terms.
onto: Take $(b_1^{-1},a_1b_1^2) \in G'$. Since $b_1^{-1} \in \mathbb{R}$ and $\ne 0$ there exists a $b_1 \in G$ such that $b_1^{-1} = \frac{1}{b_1}$, which is obviously in $G$ since it $b$ is non zero, Similarly it can be shown that $a_1b_1^2$ corresponds to $a_1,b_1 \in G$. 
Im unsure how to show how $\phi$ is order preserving however.
2) Let $G = \left\{\left.\begin{bmatrix} a & b \\ b & a \end{bmatrix}\right| \,a,b \in \mathbb{R}, a^2\ne b^2\right\}$ Let $G' = \left\{(c,d)\mid c,d \in \mathbb{R},c,a \ne 0\right\}$. Prove $G$ and $G'$ are isomorphic.
My solution:
Defining the isomorphism, we can clearly see that $\phi(G) = (a+b,a-b)$ since $c,d$ are its corresponding eigenvalues.
1-1: If we fix two points in $G'$ we can clearly see they have to correspond to the same element in $G$ since both elements are non zero.
onto: Im having a bit of trouble describing how an element in $G'$ corresponds to an element in $G$
Order-preserving: Let $(c_1,d_1), (c_2,d_2) \in G'$ since they correspond to the eigenvalues of two matrices in $G$, and $\det(AB) = \det(A)\det(B)$, we can clearly see that the operation preserves order.
3) Let G = $\left\{\left.\begin{bmatrix} 1-x & x \\-x & 1+x \end{bmatrix}\right|\, x \in \mathbb{R}\right\}$ under matrix multiplication. Prove that G is isomorphic to $\mathbb{R}$ 
4)Let $G$ and $G'$ be groups. Let $Z(G)$ and $Z(G')$ be the center of the corresponding groups. Suppouse that $\phi : G \to G'$. Prove that $\phi(Z(G)) = Z(G')$
Im stumped with the last two especially, any help would be appreciated
Thanks
 A: First, a word on intuition.  If $\phi:G\rightarrow H$ is an isomorphism, that means you should think about $G$ and $H$ as being the same group.  Saying that two groups are isomorphic is kind of like saying they are equal.  The elements may not be the same, but they work the same in every way that matters to group theorists.  (So $\#4$ for example should feel natural - they're the same group, so they should have the same center.)
Hints:


*

*Like Derek Holt said, what you're missing is a proof that $\phi(ab)=\phi(a)\phi(b)$.

*You've got the right idea, just clean up your proof.

*Map $\left(
\begin{array}{cc}
 1-x & x \\
 -x & 1+x
\end{array}
\right)$ to $x$ and prove that $$\left(
\begin{array}{cc}
 1-x & x \\
 -x & 1+x
\end{array}
\right)\left(
\begin{array}{cc}
 1-y & y \\
 -y & 1+y
\end{array}
\right)=\left(
\begin{array}{cc}
 1-x-y & x+y \\
 -x-y & 1+x+y
\end{array}
\right).$$  You can see that this is going to be isomorphic to the additive group of $\mathbb{R}$.

*If $\phi$ is a homomorphism and $ab=ba$, is it true that $\phi(ab)=\phi(ba)$?  That part just gives you $\phi(Z(G))\subseteq Z(G')$.  Bijectivity of $\phi$ gives you the reverse inclusion.
