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Find an isomorphic direct product of cyclic groups, where $V$ is an abelian group generated by $x,y,z$ and subject to relations: $$3x + 2y + 8z = 0,\qquad 2x + 4z = 0.$$

The answer is $C_{4} \oplus \mathbb{Z}$

But I don't know how to get the answers.

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Subtracting twice the second equation from the first you get $-x+2y=0$. So $x$ is superfluous. You're left with $\langle y,z : 4y+4z=0 \rangle$ which is the same as $\langle y,t : 4t=0 \rangle$, for $t=y+z$. This is $C_4 \times C_{\infty}$.

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You have a homomorphism from the free abelian group of rank $3$, generated by $x$, $y$, and $z$, onto $V$, with kernel given by the equations: $3x+2y+8z=0$, $2x+4z=0$. Finding the kernel is the same as finding the integral kernel of the matrix $$\left(\begin{array}{ccc} 3 & 2 & 8\\ 2 & 0 & 4 \end{array}\right)$$ which can be found using Gaussian elimination or Smith Normal form. $$\begin{align*} \left(\begin{array}{ccc} 3 & 2& 8\\ 2 & 0 & 4 \end{array}\right) &\to \left(\begin{array}{ccc} 1 & 2 & 4\\ 2 & 0 & 4 \end{array}\right) &&\to \left(\begin{array}{rrr} 1 & 2 & 4\\ 0 & -4 & -4 \end{array}\right)\\ &\to\left(\begin{array}{rrr} 1 & 2& 4\\ 0 & 4 & 4 \end{array}\right). \end{align*}$$ This tells you that $x+2y+4z =0$ and $4y+4z=0$. This in turn says that you can get $x$ from $y$ and $z$, that $y+z$ has order $4$, So you can generate the group with $y+z$ and $z$, subject only to the condition that $y+z$ has order $4$. This gives you the cyclic summand of order $4$ (for $\langle y+z\rangle$), and an infinite cyclic group (for $z$).

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