Question about order of elements of a subgroup Given a subgroup $H \subset \mathbb{Z}^4$, defined as the 4-tuples $(a,b,c,d)$ that satisfy $$ 8| (a-c); a+2b+3c+4d=0$$
The question is: give all orders of the elements of $\mathbb{Z}^4 /H$.
I don't have any idea how to start with this problem. Can anybody give some hints, strategies etc to solve this one?
thanks 
 A: It may be better to interpret the group $H$ as the kernel of the homomorphism:
$$\phi:\oplus_{i=1}^4\mathbb{Z}\rightarrow \mathbb{Z}_8\oplus\mathbb{Z}$$
that sends $(a,b,c,d)$ to $((a-c)\pmod8,a+2b+3c+4d)$. Verify that the kernel of $\phi$ is $H$.
I think it would be easy to determine $\text{Im}(\phi)$. Using the first isomorhpism theorem we get:
$$\oplus_{i=1}^4\mathbb{Z}/H=\oplus_{i=1}^4\mathbb{Z}/\ker(\phi)\cong \text{Im}(\phi)$$
EDIT: Since you only need to determine the possible orders of $\oplus_{i=1}^4\mathbb{Z}/H$, we can do the following:
Since $\oplus_{i=1}^4\mathbb{Z}/H\cong \text{Im}(\phi)$, therefore we only need to determine the possible orders of $Im\phi$. $Im\phi$ is a subgroup of $\mathbb{Z}_8\oplus\mathbb{Z}$. Hence the set of possible orders of $Im \phi$ is 
a subset of the set of possible orders of $\mathbb{Z}_8\oplus\mathbb{Z}$ which is $\{1,2,4,8,\infty\}$.
$\phi(0,0,0,0)$ is an element of order $1$ in $\text{Im}\,\phi$
$\phi(4,-2,0,0)$ is an element of order $2$ in $\text{Im}\,\phi$
$\phi(2,-1,0,0)$ is an element of order $4$ in $\text{Im}\,\phi$
$\phi(0,1,0,0)$  is an element of order $\infty$ in $\text{Im}\,\phi$
It is easy to show that there is no element of order $8$ in $\text{Im}\,\phi$
A: As Amir has said above, consider the homomorphism $\phi:\oplus_{i=1}^{4}\mathbb{Z}\rightarrow \mathbb{Z}_{8}\oplus\mathbb{Z}$. You can check that as Amir pointed out $\operatorname{im}(\phi)\cong (\oplus_{i=1}^{4}\mathbb{Z})/H$, so what is $\operatorname{im}(\phi)$? If you wanted, you could work this out, but you are only asked for the possible orders of elements.  In $w=(x,y)\in,\mathbb{Z}_{8}\oplus\mathbb{Z}$, then if $y\ne 0$, then $x$ has infinite order. Clearly, taking $w=\phi(0,0,0,1)$, we get $w=(0,4)$ and so \operatorname{im}(\phi)$ has elements of infinite order. 
Now consider elements of finite order. These must arise from elements of the form $w=(x,0)\in\mathbb{Z}_{8}\oplus\mathbb{Z}$. Now the possible orders of elements of $\mathbb{Z}_{8}$ are $1,2,4$ and $8$ can we find elements of such orders in $\operatorname{im}(\phi)$?
Order 1: As $\operatorname{im}(\phi)$ is a subgroup, it contains the identity - an element of order $1$.
Order 8: Consider $(a,b,c,d)\in \oplus_{i=1}^{4}\mathbb{Z}$. We want $\phi(a,b,c,d)$ to have order $8$, so $a-c$ must be coprime to $8$ (i.e. odd) and $a+2b+3c+4d=0$. Does such an element exist? Well as $a-c$ is odd, then $a$ and $c$ have different parity, so assume that $a$ is odd and $c$ is even. But then $a+2b+3c+4d$ will be off and hence not equal to $0$. Thus no elements of order $8$ exist, and in particular.
Order 4: As above, considering $\phi(a,b,c,d)$ we must have $a-c\equiv 2\text{ or }6\pmod{8}$ and $a+2b+3c+4d=0$. If $a=2$ and $c=0$, then $a-c\equiv 2\pmod{8}$. Moreover, if we then take $b=-1$ and $d=0$ we have $w=\phi(2,-1,0,0)=(2,0)$, and so $w$ is an element of order $4$.
Order 2: Take $2w=(4,0)=\phi(4,-2,0,0)$, and then $2w$ has order $2$.
Thus the possible orders of elements of $(\oplus_{i=1}^{4}\mathbb{Z})/H$ are $1,2,4$ and $\infty$.

Edit: I misread the question initially, and so thought it was asking for the possible orders of $H$ and not $\mathbb{Z}^{4}/H$. Here is an answer for finding orders of elements of $H$.
To start off with, what are the possible powers of elements of the direct product $\oplus_{i=1}^{4}\mathbb{Z}$. Clearly the only element of finite order is the identity element of order $1$ (since this is the case in $\mathbb{Z}$). Thus as $H$ is a subgroup of $\oplus_{i=1}^{4}\mathbb{Z}$, the only possible finite order of elements of $H$ is $1$. Can this order be attained?
Well clearly, the only element of $\oplus_{i=1}^{4}\mathbb{Z}$ of order $1$ is the identity element, namely $(0,0,0,0)$, and it is easy to see that this will be contained in $H$.
Now let's check if there are elements of infinite order in $H$. Suppose $x=(a,b,c,d)$ is such an element. Well for simplicity (i.e. to get rid of your first condition), just take $a=c=0$ so that $8\vert a-c=0$. Then the other condition gives $0=a+2b+3c+4d=2b+4d$, meaning that $b+2d=0$. Thus taking $d=1$ and $b=-2$ we have that $x=(0,-2,0,1)\in H$, and $x$ has infinite order.
We conclude that the possible orders of elements of $H$ are $1$ and $\infty$.
