# 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

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"Order preserving" is not what you should be trying to check. You should be deciding whether $\phi$ is a homomorphism, i.e. $\phi(gh)=\phi(g)\phi(h)$ for all $g,h \in G$, – Derek Holt Feb 24 '13 at 19:12
@DerekHolt oddly enough my book describes order-preserving as having the same property of a homomorphism and requires that we check to ensure it the same way as we would check for a homomorphism – bobdylan Feb 24 '13 at 19:21
@DerekHolt it describes as preserving order if $\phi(ab) = \phi(a)\phi(b)$ – bobdylan Feb 24 '13 at 19:22
@DerekHolt so is it the same thing? im a bit confused – bobdylan Feb 24 '13 at 19:25
@bobdylan Say you've got a homomorphism, that is, a function $\phi:G\rightarrow H$ for which $\phi(ab)=\phi(a)\phi(b)$ holds. Say that $g\in G$ has order $n$. Note that $$\phi(g^2)=\phi(g)\phi(g)=\phi(g)^2,$$ $$\phi(g^3)=\phi(g)\phi(g)\phi(g)=\phi(g)^3,$$ and so on. Then $$\phi(g^n)=\phi(g)^n=\phi(\text{id}_G)=\text{id}_H.$$ So the order of $\phi(g)$ divides $n$. – Alexander Gruber Feb 24 '13 at 19:36

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:

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

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

3. 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 $\mathbb{R}$.

4. 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.

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