# I'd like to verify that $\{, \}$ is an abelian group under multiplication

Since $ =  \in \mathbb Z_2$, this ring is closed under multiplication.

Let $$ be identity: $ = $ and $ = $.

$ = [0 \cdot 1] = [1 \cdot 0] = $ so commutativity holds.

If we assume $\mathbb Z_2$ is a ring, associativity is given.

I am having difficulty showing there's inverse element in $\mathbb Z_2$. What could possibly be the inverse of $$? Please, elaborate on this. Thanks.

• $=$ and $=$. Since $\mathbb{Z}_2=\{,\}$, we conclude that $\forall x\in\mathbb{Z}_2,\ x=$. Since $\neq$, we conclude that $\forall x\in\mathbb{Z}_2,\ x\neq$. Doesn't this show that $$ has no inverse? – gniourf_gniourf Feb 16 '16 at 18:18
• The additive zero does not have a multiplicative inverse in a ring. – John Smith Feb 16 '16 at 18:19
• In a field, the non-zero elements form a group under multiplication. (The zero element in a ring only has an inverse in the trivial ring with only one element.) – Rob Arthan Feb 16 '16 at 18:22
• You need to reread the definition of a field. – Tobias Kildetoft Feb 16 '16 at 18:23
• $(\mathbb{Z_2},+)$ isn't the same as $(\mathbb{Z_2^{*}},(*))$ – user111750 Feb 16 '16 at 18:25