Came across an interesting question regarding subgroups of $\mathbb{Z}$. Let $A \subseteq \mathbb{Z}$ such that $0 \in A$, $A = -A$ (for every element in $A$, its negative is also in $A$), and $A + 2A \subseteq A$ (for every $a,b \in A $, $a+2b \in A$). We need to prove that this implies $A$ must be a subgroup of $\mathbb{Z}$. Moreover, we need to show that this does not necessarily hold for $A \subseteq \mathbb{Z}^2 $ which has the same properties.
Proving subgroups seems to be a challenge, since you get associativity, identity, and inverse for free. However, you cannot prove closure by manipulating the elements (atleast I cannot). I think the solution lies in the second part of the answer, where $A$ has some property that only exists as a subset of $\mathbb{Z}$ and not $\mathbb{Z}^2$. Any help would be appreciated.
Best Answer
First consider the case $A\subseteq \mathbb{Z}$ . . .
If $A=\{0\}$, then $A$ is the trivial subgroup of $\mathbb{Z}$.
Suppose $A\ne\{0\}$.
Since $A$ is closed under negation, $A$ must have a least positive element, $a$ say.
Claim:$\;A=\langle{a}\rangle\;$(the cyclic subgroup of $\mathbb{Z}$ generated by $a$).
Proof:
By an easy induction, $0+2ka\in A$ for all nonnegative integers $k$.
Thus if $n$ is an even nonnegative integer, then $n=2k$ for some nonnegative integer $k$, hence $na=2ka=0+2ka\in A$.
Similarly, by an easy induction, $a+2ka\in A$ for all nonnegative integers $k$.
Thus if $n$ is an odd positive integer, then $n=2k+1$ for some nonnegative integer $k$, hence $na=(2k+1)a=a+2ka\in A$.
Combining both cases, and noting that $A$ is closed under negation, it follows $na\in A$ for all integers $n$.
Hence $\langle{a}\rangle\subseteq A$.
To show $A=\langle{a}\rangle$, suppose instead that we have the proper inclusion $\langle{a}\rangle\subset A$.
Then $A{\setminus}\langle{a}\rangle\ne{\large{\varnothing}}$.
Our goal is to derive a contradiction.
Since $A$ is closed under negation, and $0\in A$, it follows that $A{\setminus}\langle{a}\rangle$ has a least positive element, $b$ say.
By minimality of $a$, we must have $b > a$.
By hypothesis $b-2a\in A$ and since $b\not\in\langle{a}\rangle$, it follows that $b-2a\in A{\setminus}\langle{a}\rangle$.
Since $b-2a < b$, the minimality of $b$ implies $b-2a < 0$.
Thus, $a < b < 2a$.
By hypothesis $2a-b\in A$ and since $b\not\in\langle{a}\rangle$, it follows that $2a-b\in A{\setminus}\langle{a}\rangle$.
But from $a < b < 2a$, we get $0 < 2a - b < b$, contrary to the minimality of $b$.
This completes the proof.
Next consider the case $A\subseteq \mathbb{Z}^2$ . . .
To show that $A$ need not be a subgroup of $\mathbb{Z}^2$, consider the set $$A=\{(x,y)\in\mathbb{Z}^2\mid\;\text{at least one of $x,y$ is even}\}$$ Then $A$ satisfies the hypothesis, but $A$ is not closed under addition since $(1,0)\in A$ and $(0,1)\in A$, but $(1,1)\not\in A$.