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G is the set of matrices of the form $G=$$\begin{pmatrix} x & x \\ x & x \end{pmatrix}$. So for this set to be a group I know it needs to be:

  • Closed under matrix multiplication
  • The Associative Property holds
  • Contains an Identity Element
  • Every element needs to have an inverse

So the form of the matrices is such that all the elements are the same but not 0. How do I go about proving these?

Working through this problem, I seem to have hit a contradiction. Since G is a subgroup of the bigger $2x2$ nonsingular matrices group why does G not have the same identity element as its parent group? Namely \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}

Isn't the subgroup supposed to have the same identity element as its parent group?

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    $\begingroup$ For starters, what do you expect the identity element to look like? $\endgroup$ – Andrew D. Hwang Feb 14 '14 at 1:56
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    $\begingroup$ But these matrices are not invertible. $\endgroup$ – user99680 Feb 14 '14 at 1:57
  • $\begingroup$ @user99680 Since the identity of the group is NOT the identity matrix, it doesn't matter. $\endgroup$ – N. S. Feb 14 '14 at 2:00
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    $\begingroup$ @N.S: But under the standard multiplication, these matrices are not invertible.Ah, maybe there is an identity for the subgroup that is not the same as the identity for the group? $\endgroup$ – user99680 Feb 14 '14 at 2:07
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    $\begingroup$ @user99680 Yes, but it doesn't matter. It only shows it is not a subgroup of $GL_2(\mathbb R)$. But there are groups of non-invertible matrices. It simply means that the identity of the group is not the identity matrix.... A simpler example is the following: matrices of the type $\begin{bmatrix} a & 0 \\ 0 & 0 \end{bmatrix}$ with $a \neq 0$ are a group under multiplication, even if they are not invertible. The identity in this case is $\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$ and the inverse is $\begin{bmatrix} \frac1{a} & 0 \\ 0 & 0 \end{bmatrix}$...And when you multiply these matrices $\endgroup$ – N. S. Feb 14 '14 at 2:14
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Hint

1) What is $\begin{bmatrix} x & x \\ x & x \end{bmatrix}\begin{bmatrix} y & y \\ y & y \end{bmatrix}$?

2) The multiplication of matrices is associative.

3) When you are looking for the identity you want

$$\begin{bmatrix} x & x \\ x & x \end{bmatrix}\begin{bmatrix} e & e \\ e & e \end{bmatrix}=\begin{bmatrix} x & x \\ x & x \end{bmatrix}$$

Now, do the multiplication on the left, what do you get?

4) With the $e$ from $3)$ solve

$$\begin{bmatrix} x & x \\ x & x \end{bmatrix}\begin{bmatrix} y & y \\ y & y \end{bmatrix}=\begin{bmatrix} e & e \\ e & e \end{bmatrix}$$

for $y$. Again, all you need to do is doing the multiplication...

P.S. In order for this to be a group, you need $x \neq 0$.

P.P.S Since $\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}=2\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$, you can prove that

$$F: \mathbb R \backslash\{0 \} \to G$$ $$F(x) =\frac{x}{2} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$$

is a bijection and it preserves multiplications. Since $\mathbb R \backslash\{0 \}$ is a group it follows that G must also be a group and $F$ is an isomorphism... But this is probably beyond what you covered so far...

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  • $\begingroup$ 1. $\begin{matrix} 2xy & 2xy \\ 2xy & 2xy \end{matrix}$ 2. Yup thats the easy one. 3. You did the multiplication on the left? Multiplying x by the identity element will give you x, like you have above.. $\endgroup$ – atherton Feb 14 '14 at 2:15
  • $\begingroup$ @RoyM. The multiplication on the left is $\begin{bmatrix} 2xe & 2xe \\ 2xe & 2xe \end{bmatrix}$. Now for which value of $e$ do you get:$$\begin{bmatrix} 2xe & 2xe \\ 2xe & 2xe \end{bmatrix}=\begin{bmatrix} x & x \\ x & x \end{bmatrix} \, ? ;)$$ $\endgroup$ – N. S. Feb 14 '14 at 2:18
  • $\begingroup$ Are you talking about your first bullet? \begin{bmatrix} x & x \\ x & x \end{bmatrix}\begin{bmatrix} y & y \\ y & y \end{bmatrix} For which value of $e$ do I get what? I am sorry I am not following! $\endgroup$ – atherton Feb 14 '14 at 3:01
  • $\begingroup$ @RoyM. I am speaking about $3$. Keep in mind that identity in a group of matrices doesn't necessarily mean identity matrix ;) $\endgroup$ – N. S. Feb 14 '14 at 3:05
  • $\begingroup$ Ahh I see what you're saying. And for your first comment, $e=\frac{1}{2}$? $\endgroup$ – atherton Feb 14 '14 at 3:17

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