# Study of Affine Group

Affine group G:= $Aff_{n} (R)$ where n is a positive integer and $Aff_{n} (R)$ is the subset in $GL_{n+1}(R)$  such that:

$Aff_{n} (R)$ := $\left \{ \begin{pmatrix} a & b \\ 0 & 1 \end{pmatrix} \right \}$  where $a ? GL_{n}(R), b ? M_{n,1}(R)$
$M_{n,1}(R)$ denotes the set of $n\times 1$ matrices, 0 denotes the $1\times n$  zero matrix, 1 denotes the $1\times 1$ matrix.

$M$ := $\left \{ \begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix} \right \}$ where $b ? M_{n,1}(R)$
$N$ := $\left \{ \begin{pmatrix} a & 0 \\ 0 & 1 \end{pmatrix} \right \}$ where where $a ? GL_{n}(R)$

Questions:
1. Show that $M$ is a subgroup of $GL_{n+1}(R)$ is isomorhic to the vector space $R^n$ under addition as a group.
2. Show that $N$ is a subgroup of $GL_{n+1}(R)$ isomorphic to $GL_{n}(R)$
3. Show that $Aff_{n} (R)$= $MN$.

*I don't need the answers to be in details, just short and simple answers would be fine enough.

• The short and simple answer would be "do the products and see that they behave in the right way", if you want the actual computation to be shown then it takes a while to answer.

• Also I assume the top-left 1 in M denotes the nxn identity matrix.

• Yes it is the non identity matrix and the top-right 0 in N is zero matrix

• I know how to answer your question, but honestly, I don't see any intermediate ground between saying "it's immediate if you do the computation" and actually doing the computation. The former, hey, I just gave it to you for free; for the latter, 4$aren't nearly worth the time. All you have to do is a check. • Are you willing to help me with these questions? I really need your help, please. • I am sorry, but I already answered a question of yours that was priced well below the actual value of the question, I cannot do that indefinitely. Properly answering this question will take a lot of time, I could take both for about 30$ total but definitely not 10$. If you can't or don't want to spend that much, I'm telling you that all you have to do to solve it yourself is to actually try multiplying two elements of M, two of N, and one of each, and see what you get. • Your offers are low for the level of your questions with such early deadlines. • How about 15$ for both questions? I am sorry but I really not able to pay so much.

• I didn't even notice the deadline, but yes, David89 is right. It's a long question to be solved urgently on a Saturday night. Think about how much would it cost to have a private teacher urgently solve these for you at night during weekend.

• Still a no for me. I understand if you can't or don't want to raise the amount, but I'm not taking these for less. Maybe someone else will, but I think it's unlikely, and that your best option is to try to do what I previously suggested. In any case, best of luck.

• Are you ok with $30 for both questions? • It's on the low end, but I said that would be okay before and I'm not taking that back. Yes, I'll take them at 30$ for both.

Answers can be viewed only if
1. The questioner was satisfied and accepted the answer, or
2. The answer was disputed, but the judge evaluated it as 100% correct.
• To prove that M and N are subgroups, I need to show that they contain the identity, but that's easy, just take A=I for M and b=0 for N.

• May I know the meaning of G in the answer? Is it the affine group?

• Yes, didn't you define G = Aff_n(R) at the beginning of the question?

• Sorry I overlooked it. Is subgroup of G similar to subgroup of GL_n+1(R)? because the question requires subgroup of GL_n+1(R)

• I guess there is something wrong with your solution. I think the matrix (A 0;0 1) should be ∈N but not ∈M?

• G is a subgroup of GL_(n+1)(R), so subgroup of G implies subgroup of GL.

• I don't understand why that would mean that there is something wrong. What's the problem?

• In the question, N is defined as matrix (A 0;0 1) but in the solution you wrote (A 0;0 1) is the element in M, which I think should be N. I'm sorry if I misunderstood this.

• Oh, you are right, there's a typo, I'm sorry. In the last section you should swap M and N. It's essentially the same, but you have to swap the terms of the product (which makes it easier): you have that (I b; 0 1)(A 0; 0 1) = (A, b; 0, 1). Basically, replace A^{-1} b with 0 and swap the terms, and the result is the same. Do you understand or you need further clarification?

• It's ok I can understand it. Thank you for your clarification :)