# How should i prove this homomorphism? [closed]

Show that the map $x:M\rightarrow \mathbb{R}$, $\mathbb{R}$ has group structures under addition, is defined by $x\left( \left[ \begin{matrix} a& b\\ o& c\end{matrix} \right] \right)$ =$\log \left( \dfrac {a} {c}\right)$ is a group homomorphism.

Also, is x surjective and what is ker(x)?

## closed as off-topic by Ken Duna, uniquesolution, C. Falcon, Alex Provost, Daniel W. FarlowMar 25 '17 at 22:24

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Ken Duna, uniquesolution, C. Falcon, Alex Provost, Daniel W. Farlow
If this question can be reworded to fit the rules in the help center, please edit the question.

• As Alan points out below, it helps us if you provide more detail in your questions. Defining $M$ in this problem would be helpful. However, I would like even more to see you tell us what your thoughts on the problem are. For instance, what have you tried to do? What is your background in group theory like? These things help us understand you as a blossoming mathematician which in turn helps us to know how best to teach and instruct you. It is the lack of these sorts of things in your post that caused your question to be flagged for closure. – Ken Duna Mar 25 '17 at 14:25

I assume that $M \subseteq GL(2,\Bbb{R}).$ \begin{align*} x\left( \begin{bmatrix} a& b\\ 0& c\end{bmatrix} \begin{bmatrix} p& q\\ 0& r\end{bmatrix}\right) &= x\left( \begin{bmatrix} ap& aq+br\\ 0& cr\end{bmatrix}\right)\\ &= \log \left( \dfrac {ap} {cr}\right) \\ &=\log\left(\frac{a}{c}\right)+\log\left(\frac{p}{r}\right)\\ &= x\left( \begin{bmatrix} a& b\\ 0& c\end{bmatrix}\right)+x\left( \begin{bmatrix} p& q\\ 0& r\end{bmatrix}\right) \end{align*}

Beware that the operation of $M$ is multiplication.

The kernel of $x$ is $\left\{\begin{bmatrix} a& b\\ 0& c\end{bmatrix}\; \bigg|\;\log(a/c)=0 \right\}=\left\{\begin{bmatrix} a& b\\ 0& a\end{bmatrix}\; \bigg|\;a,b\in \Bbb{R},a\neq0\right\}$.

For $k\in \Bbb{R}$, $$x\left( \begin{bmatrix} e^k& 0\\ 0& 1\end{bmatrix} \right) =\log(e^k)=k$$ So the map is surjective.

• @Anastasia I have visited your profile and have a look on the previous questions you had asked. Try to add more details or your idea about the question. Also accept the answer if you think that the answer solved your problem. For example in this question your $M$ and its operation is not explained. – Alan Wang Mar 25 '17 at 14:11
• I haven't got used to it, but i'll pay attention to your advices. – Anastasia Mar 25 '17 at 14:22

I'll assume that $M \subset \mathrm{GL}_2(\mathbb{R})$ , where all of the elements in $M$ are upper triangular matrices.

To show that $x$ is a homomorphism, we just need to show that it preserves the identity and the operations, i.e. $x\left( I \right)= 0$ and for any $a,b \in M$, $x(ab)=x(a)+x(b)$.

Alan Wang did a nice job showing the later, and it's clear that $x \left( \begin{pmatrix} 1& 0 \\ 0 & 1 \end{pmatrix} \right) = \log(1/1) = 0$, however, I disagree on his kernal.

$\ker(x)$ should be all the elements that get sent to 0, i.e. $$\ker(x) = \Bigg \{ \begin{pmatrix} a & b \\ 0 & a \end{pmatrix} \mid a,b \in \mathbb{R} \Bigg \}$$

(since $\log \left(\dfrac{a}{a} \right)= 0$ ).

$x$ is also surjective, Alan did a great job showing that as well.

PS: I would have just added this as a comment to the previous answer, but it turns out I don't have enough reputation to do so :(

Edit: I had the kernal wrong

• $\log(\frac{a}{a^{-1}}) = \log(a^2) = 2\log(a)$... – Ken Duna Mar 25 '17 at 14:28
• You're soooo right, I was thinking about the determinant for some reason lol, will remove my sorry answer – Yahya Fidouh Mar 25 '17 at 14:38
• $\log(a/c) = 1$ is still wrong I think. – Yahya Fidouh Mar 25 '17 at 14:41
• You are right. I can't believe I missed that when I was editting. – Ken Duna Mar 25 '17 at 14:43