Homomorphism

Jan 2020
14
0
UK
Have to use letters. p and q have to represent ANY element of P. If you use examples you can show it, sure, but all that means is that the proof holds for those examples. It means nothing for the entire group.

Also, use the function symbol given to you so you avoid confusing the grader, if there is one. φ
Got it. Also I proved in part a that P was a group so does that cover point one enough? Or do I have to do it again just for addition
 
Oct 2018
129
96
USA
A group has to have some sort of operation paired with it. On second look, $P$ may not form a group under addition due to the constraint $dg-ef \neq 0$. When you prove closure, you’ll have to also show that the result of performing the group operation also has that $dg-ef \neq 0$ quality.
 
Jan 2020
14
0
UK
Right I think I’ve done the first 2 parts but you know the bit that says find the ker(thigh) how would I do that bit?
 
Oct 2018
129
96
USA
The Kernel in this case is what elements from $P$ map to the $0$ element of $GL$. Be careful, though. Since all elements of $GL$ are invertible, the $0$ Matrix isn’t in $GL$. You need to find out what the null element of $GL$ is.
 
Jan 2020
14
0
UK
The Kernel in this case is what elements from $P$ map to the $0$ element of $GL$. Be careful, though. Since all elements of $GL$ are invertible, the $0$ Matrix isn’t in $GL$. You need to find out what the null element of $GL$ is.
How do I do that? I’m very knew to this terminology. Is there a set method
 
Oct 2018
129
96
USA
I think it might be easier for me to just go through this problem and you ask any questions you have about what I've done.

First, we need to show that $P$ is a group under multiplication. For this to be true $P$ has to satisfy the 4 group axioms.

1. Closure -------

Given 2 elements of $X,Y \in P$ we have to show that $X \cdot Y \in P$

$
\begin{pmatrix}
x_1 & x_2 & x_3\\
0 & x_4 & x_5 \\
0 & x_6 & x_7
\end{pmatrix}

\begin{pmatrix}
y_1 & y_2 & y_3\\
0 & y_4 & y_5 \\
0 & y_6 & y_7
\end{pmatrix}

=

\begin{pmatrix}
x_1 y_1 & * & * \\
0 & x_4 y_4 + y_5 y_6 & x_4 y_5 + x_5 y_7 \\
0 & x_6 y_4 + x_7 y_6 & x_6 y_5 + x_7y_7
\end{pmatrix}
=XY
$

For $XY$ to be in $P$ we have to show that

$\text{det}

\begin{pmatrix}
x_4 y_4 + y_5 y_6 & x_4 y_5 + x_5 y_7 \\
x_6 y_4 + x_7 y_6 & x_6 y_5 + x_7y_7
\end{pmatrix}
\neq 0
$

$\text{det}

\begin{pmatrix}
x_4 y_4 + y_5 y_6 & x_4 y_5 + x_5 y_7 \\
x_6 y_4 + x_7 y_6 & x_6 y_5 + x_7y_7
\end{pmatrix}
= (x_4 y_4 + y_5 y_6)(x_6 y_5 + x_7y_7) - (x_4 y_5 + x_5 y_7)(x_6 y_4 + x_7 y_6)
$

After multiplying this out and pulling like terms you'll find that

$
(x_4 y_4 + y_5 y_6)(x_6 y_5 + x_7y_7) - (x_4 y_5 + x_5 y_7)(x_6 y_4 + x_7 y_6) = (x_4x_7 - x_5x_6)(y_4y_7-y_5y_6) =
$

$
\text{det}
\begin{pmatrix}
x_4 & x_5 \\
x_6 & x_7
\end{pmatrix}
\text{det}
\begin{pmatrix}
y_4 & y_5 \\
y_6 & y_7
\end{pmatrix}
$

$X$ and $Y$ are in $P$, so by definition these determinants are nonzero, so their product is nonzero.

Since both $X$ and $Y$ are in $P$, $x_1 \neq 0$ and $y_1 \neq 0$ so $x_1 y_1 \neq 0$. Fields are closed under addition and multiplication so all the elements of $XY$ are in $\mathbb{F}$.

Lastly, since

$\text{det}

\begin{pmatrix}
x_4 y_4 + y_5 y_6 & x_4 y_5 + x_5 y_7 \\
x_6 y_4 + x_7 y_6 & x_6 y_5 + x_7y_7
\end{pmatrix}

=

\text{det}
\begin{pmatrix}
x_4 & x_5 \\
x_6 & x_7
\end{pmatrix}
\text{det}
\begin{pmatrix}
y_4 & y_5 \\
y_6 & y_7
\end{pmatrix} \neq 0

$

We have that $XY$ must also be in $P$



2. Associativity ----------

Inherent in matrix multiplication.



3. Identity --------------

The identity in matrix multiplication,

$
\begin{pmatrix}
1 & 0 & 0\\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}
$

Is in $P$, I'll leave that to you to check.



4. Inverse ------------------

We need to show every element of $P$ has a multiplicative inverse.

For a matrix to have an inverse, it needs to be invertible, we means it's determinant is nonzero. We'll use $X$ as an arbitrary element of $P$ again.

$\text{det}

\begin{pmatrix}
x_1 & x_2 & x_3\\
0 & x_4 & x_5 \\
0 & x_6 & x_7
\end{pmatrix}

=

x_1 \text{det}
\begin{pmatrix}
x_4 & x_5 \\
x_6 & x_7
\end{pmatrix}
-0 + 0
$

Since $x_1 \neq 0$ and that determinant is nonzero, all the elements of $P$ are invertible.


Since all 4 axioms are satisfied, $P$ is a group under matrix multiplication.
 
Last edited:
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Oct 2018
129
96
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Now, for the homomorphism part. We'll use the same $X$ and $Y$ again. We have to show that

$\varphi(XY) = \varphi(X) \varphi(Y)$

We know this is the same as

$
\begin{pmatrix}
x_4 y_4 + y_5 y_6 & x_4 y_5 + x_5 y_7 \\
x_6 y_4 + x_7 y_6 & x_6 y_5 + x_7y_7
\end{pmatrix}
=

\begin{pmatrix}
x_4 & x_5 \\
x_6 & x_7
\end{pmatrix}

\begin{pmatrix}
y_4 & y_5 \\
y_6 & y_7
\end{pmatrix}
$

Which is true. So $\varphi$ is a group homomorphism.


Now we show $\varphi$ is onto.

For any element of $G \in \text{GL}(2, \mathbb{F})$ we have that (where $a \neq 0$)

$
p=
\begin{pmatrix}
a & 0 & 0\\
0 & g_1 & g_2\\
0 & g_3 & g_4
\end{pmatrix}

\mapsto^{\varphi}

\begin{pmatrix}
g_1 & g_2\\
g_3 & g_4
\end{pmatrix}
=G
$

Since every element of $\text{GL}(2, \mathbb{F})$ is mapped to, $\varphi$ is onto.


Lastly, the kernel. The kernel of a group homomorphism is defined to be the set of all elements of the domain group that map to the identity element of the codomain group. In this case, the elements of $P$ that map to the identity matrix. We have that for any $x_1 \neq 0$ and $x_1, x_2 , x_3 \in \mathbb{F}$,

$
\begin{pmatrix}
x_1 & x_2 & x_3 \\
0 & 1 & 0\\
0 & 0 & 1
\end{pmatrix}

\mapsto^{\varphi}

\begin{pmatrix}
1 & 0\\
0 & 1
\end{pmatrix}
$

Formally:

$\text{ker}(\varphi) = \left\{

\begin{pmatrix}
x_1 & x_2 & x_3 \\
0 & 1 & 0\\
0 & 0 & 1
\end{pmatrix}

: x_1 , x_2 , x_3 \in \mathbb{F} , x_1 \neq 0 \right\}
$
 
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Jan 2020
14
0
UK
Now, for the homomorphism part. We'll use the same $X$ and $Y$ again. We have to show that

$\varphi(XY) = \varphi(X) \varphi(Y)$

We know this is the same as

$
\begin{pmatrix}
x_4 y_4 + y_5 y_6 & x_4 y_5 + x_5 y_7 \\
x_6 y_4 + x_7 y_6 & x_6 y_5 + x_7y_7
\end{pmatrix}
=

\begin{pmatrix}
x_4 & x_5 \\
x_6 & x_7
\end{pmatrix}

\begin{pmatrix}
y_4 & y_5 \\
y_6 & y_7
\end{pmatrix}
$

Which is true. So $\varphi$ is a group homomorphism.


Now we show $\varphi$ is onto.

For any element of $G \in \text{GL}(2, \mathbb{F})$ we have that (where $a \neq 0$)

$
p=
\begin{pmatrix}
a & 0 & 0\\
0 & g_1 & g_2\\
0 & g_3 & g_4
\end{pmatrix}

\mapsto^{\varphi}

\begin{pmatrix}
g_1 & g_2\\
g_3 & g_4
\end{pmatrix}
=G
$

Since every element of $\text{GL}(2, \mathbb{F})$ is mapped to, $\varphi$ is onto.


Lastly, the kernel. The kernel of a group homomorphism is defined to be the set of all elements of the domain group that map to the identity element of the codomain group. In this case, the elements of $P$ that map to the identity matrix. We have that for any $x_1 \neq 0$ and $x_1, x_2 , x_3 \in \mathbb{F}$,

$
\begin{pmatrix}
x_1 & x_2 & x_3 \\
0 & 1 & 0\\
0 & 0 & 1
\end{pmatrix}

\mapsto^{\varphi}

\begin{pmatrix}
1 & 0\\
0 & 1
\end{pmatrix}
$

Formally:

$\text{ker}(\varphi) = \left\{

\begin{pmatrix}
x_1 & x_2 & x_3 \\
0 & 1 & 0\\
0 & 0 & 1
\end{pmatrix}

: x_1 , x_2 , x_3 \in \mathbb{F} , x_1 \neq 0 \right\}
$
Now, for the homomorphism part. We'll use the same $X$ and $Y$ again. We have to show that

$\varphi(XY) = \varphi(X) \varphi(Y)$

We know this is the same as

$
\begin{pmatrix}
x_4 y_4 + y_5 y_6 & x_4 y_5 + x_5 y_7 \\
x_6 y_4 + x_7 y_6 & x_6 y_5 + x_7y_7
\end{pmatrix}
=

\begin{pmatrix}
x_4 & x_5 \\
x_6 & x_7
\end{pmatrix}

\begin{pmatrix}
y_4 & y_5 \\
y_6 & y_7
\end{pmatrix}
$

Which is true. So $\varphi$ is a group homomorphism.


Now we show $\varphi$ is onto.

For any element of $G \in \text{GL}(2, \mathbb{F})$ we have that (where $a \neq 0$)

$
p=
\begin{pmatrix}
a & 0 & 0\\
0 & g_1 & g_2\\
0 & g_3 & g_4
\end{pmatrix}

\mapsto^{\varphi}

\begin{pmatrix}
g_1 & g_2\\
g_3 & g_4
\end{pmatrix}
=G
$

Since every element of $\text{GL}(2, \mathbb{F})$ is mapped to, $\varphi$ is onto.


Lastly, the kernel. The kernel of a group homomorphism is defined to be the set of all elements of the domain group that map to the identity element of the codomain group. In this case, the elements of $P$ that map to the identity matrix. We have that for any $x_1 \neq 0$ and $x_1, x_2 , x_3 \in \mathbb{F}$,

$
\begin{pmatrix}
x_1 & x_2 & x_3 \\
0 & 1 & 0\\
0 & 0 & 1
\end{pmatrix}

\mapsto^{\varphi}

\begin{pmatrix}
1 & 0\\
0 & 1
\end{pmatrix}
$

Formally:

$\text{ker}(\varphi) = \left\{

\begin{pmatrix}
x_1 & x_2 & x_3 \\
0 & 1 & 0\\
0 & 0 & 1
\end{pmatrix}

: x_1 , x_2 , x_3 \in \mathbb{F} , x_1 \neq 0 \right\}
$
Thank you so much! This makes a lot more sense now to me