I am working on the following problem to prepare for an exam.
In the following cases, find a basis {$α_1$, ..., $α_n$} for $G$ such that the subgroup $H$ has a basis of the form {$m_1·α_1$, ..., $m_k ·α_k$} for some integers $m_1, ..., m_k$.
(1) $G = Z^2$ , H is all $(a, b)$ such that $a + b$ is even.
(2) $G = Z^3$ , $H$ is all $(a, b, c)$ such that 7 divides $a + b + c$.
(3) $G = Z^4$ , $H$ is all $(a, b, c, d)$ such that $a + b = 0$ and $c + d = 0$.
Here is what I have thought about so far.
For $1$, I know that if $a + b$ is even then the elements must have the form $(2k, 2j)$ or $(2k + 1, 2j + 1)$. I believe a basis for $H$ would then be $(2,0),(1,1)$. However, I know that a basis for $Z^2$ must have length $2$ so I believe that $(1,1), (0,1)$ satisfies this since this definitely "spans" $Z^2$.
For $2$ I have thought that since $7 | a + b +c$ then $H$ must be generated by $(7,0,0), (0,7,0), (5,1,1)$ (I'm not entirely sure if this is correct, though) and so the basis for $Z^3$ would be $(1,0,0),(0,1,0),(5,1,1)$ which is indeed a basis.
Lastly, for $3$ I have thought that if $a + b = 0$ then we must have one of the spanning vectors be $(1,-1,0,0)$ and the other spanning vector would be of the form $(0,0,1,-1)$. For any element $h \in H$, $h = (a,-a,c,-c) = a(1,-1,0,0) + b(0,0,1,-1)$ so this is indeed a basis for $H$. So extending this to a basis for $G$, yields for example $(1,-1,0,0), (0,0,1,-1), (0,1,0,0), (0,0,1,0)$.
I would greatly appreciate it if someone could correct, and perhaps let me know if there is a simpler way to do this kind of problem.