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,I'm trying to solve this problem.

I have two additive groups $G, H$. The first is the group of matrices $4x1$ with coefficients in $\Bbb Z_{11}$. The second is the group of matrices $3x1$ with coefficients in $\Bbb Z_{11}$.

Let a = $\left(\begin{matrix} 2 & 7 & 2 & 4 \\ 3 & 5 & 1 & 2 \\ 5 & 1 & 8 & 5 \end{matrix}\right)$ (coefficients in $\Bbb Z_{11}$) and $f: G \rightarrow H$ defined as $x \rightarrow a*x$.


Prove that $f$ is a homomorphism of additive groups. Tell which is the kernel and its order.

Is $\left(\begin{matrix} 4 \\ 6 \\ 10 \end{matrix}\right)$ $\in H$ in the image of $f$?

Since it is a very long and mechanical approach, I don't write my attempt to try that the function is a homomorphism (it is, maybe someone could confirm). Now I'm trying to find its kernel. I reasoned in a similar way to the approach used to verify the first point.

First of all, I consider a general matrix $\in G$: $\left(\begin{matrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{matrix}\right)$ then I need to do the multiplication a$*$b. I obtain the matrix:

$\left(\begin{matrix} 2x_1+7x_2+2x_3+4x_4 \\ 3x_1+5x_2+x_3+2x_4 \\ 5x_1+1x_2+8x_3+5x_4 \end{matrix}\right)$

Now, by the kernel definition, I need to find when for certain matrices $\in G$:$\left(\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}\right)$

It is sufficient to solve a system of three linear equations? In particular: $2x_1+7x_2+2x_3+4x_4=0 \\ 3x_1+5x_2+x_3+2x_4=0\\ 5x_1+1x_2+8x_3+5x_4=0$

If this is a good approach, how do I need to continue after I solved the system? How can I find the order of the kernel? Can I apply the same approach to answer the last question? If yes, which tells me that the element belongs or not the image of the function?

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  • $\begingroup$ Your approach is good, Once you solve the system, you will express it as a combination of matrices in $G$. If done correctly, then you will have linearly independent vectors that will form a basis of the kernel. With the same system and right hand side vector changed appropriately you can check if the given vector is in the range, $\endgroup$ – Anurag A Mar 14 at 19:12
  • $\begingroup$ @AnuragA Hi, I tried to solve the system. I obtained the following matrix (assuming I didn't make a mistake in calculations) $\left(\begin{matrix} 1 & 9 & 0 & 0 &0 \\ 0 & 0 & 1 & 0 &0\\ 0 & 0 & 0 & 1 &0 \end{matrix}\right)$ so let $x_2=t, t\in \Bbb Z_{11}$ the solutions are $x_1=-9t, x_2=t, x_3=0, x4=0$. Now how I need to continue? $\endgroup$ – PCNF Mar 15 at 17:04
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Edited: I didn't realize that the last matrix you had in your comments is an augmented matrix, hence I had added $x_5$.

So you should have $$\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}=t\begin{bmatrix}-9\\1\\0\\0\end{bmatrix}, \quad t \in \mathbb{Z}_{11}.$$

Thus $$\ker(f)=\left\{t\begin{bmatrix}-9\\1\\0\\0\end{bmatrix} \quad | \, \quad t \in \mathbb{Z}_{11}\right\}.$$ This is a cyclic subgroup generated by the element $\begin{bmatrix}2\\1\\0\\0\end{bmatrix} $ whose order is $11$ (since both $2$ and $1$ have orders $11$.)

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  • $\begingroup$ Thanks for details. Why $x_5$? The system has only 4 unknowns... $\endgroup$ – PCNF Mar 15 at 19:37
  • $\begingroup$ @PCNF I have edited the answer. I mistook your last matrix in your comment as the actual matrix. $\endgroup$ – Anurag A Mar 16 at 13:29

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