Let $A $ be a subset of an abelian group $H$. Then if $G = \{ a + b : a, b \in A\}$ is a group and $A$ is closed under taking negatives, then is $A$ also a group?

  • 1
    $\begingroup$ Have you tried an examples? For instance, what about if $H$ is the group of order $2$, where there's only four subsets to test? $\endgroup$ Jan 13, 2016 at 20:00
  • $\begingroup$ A group $H\subset G$ is a subgroup of $G$ if and only if $a,b\in H\rightarrow ab^{-1}\in H$ $\endgroup$
    – Peter
    Jan 13, 2016 at 20:00

2 Answers 2


Probably the simplest counterexample is given by $G=(\mathbb{Z},+)$, with $A$ the set of odd numbers. More generally, if $G$ has a subgroup $H$ of index $2$, then $A=G\setminus H$ will do.

  • $\begingroup$ So essentially you notice that [with $H=(\mathbb{Z},+)$] $G=2\mathbb{Z}$ when $A$ is the set of odd numbers. While $2\mathbb{Z}$ is a subgroup of $H$, $A$ however isn't. This seems to work as an example when ever you take $A=H\setminus K$ where $K$ is a normal subgroup with index 2. $\endgroup$
    – user160110
    Jan 14, 2016 at 18:00

Since A is closed, we have

$a+0=a$, Since $a\in A$ due to closure property, $0\in A$,hence existence of identity.

$a+a^{-1}=0$. $0\in $A, due to closure property, $a^{-1}\in A$. Hence existence of inverse.

Also check for associative property. Let $c\in A$



Since $c+(a+b)=(c+a)+b$, Associative property holds.

Hence A must be a group.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .