Suppose H is a subring of R such that for every $r∈H$ we have $r^2=e$. Prove that $H$ is commutative.

Question

So I know that since H is a subring of $\mathbb{R}$ that H is then by definition:

1) closed under addition

2) closed under multiplication

3) have the same multiplicative identity as $\mathbb{R}$

Am I missing anything? I think I need to only reference fact 2, which states that H is closed under multiplication for this proof.

I tried looking up on the internet how to show that a subring is commutative and did not find anything. Is the answer as simple as saying that since $\mathbb{R}$ is commutative then the subring H, is itself by definition commutative?

Or do I go about it by showing that lets say for example:

Let $r_1, r_2 \in H$ then $r_1 \times r_2 = e = r_1 \times r_2$ which is then commutative?

Answer 1

Let us assume that every $0 \ne r \in H$ satisfies $r^2 = e$, ruling out the case $e = 0^2 = 0$, which trivializes everything. These words inspired by the comment of Rob Arthan to the question itself.

Let

$x, y \in H; \tag 0$

then

$x^2 = y^2 = e; \tag 1$

also,

$xy \in H, \tag{1.5}$

whence

$xyxy = (xy)^2 = e; \tag 2$

then

$yxy = eyxy = x^2yxy = x(xyxy) = xe = x, \tag 3$

and

$yx = yxe = yxy^2 = (yxy)y = xy, \tag 4$

which shows $H$ is commutative. $OE\Delta$.

Answer 2

Commutativity means that for all $r,s\in H$ we have that $r\circ s= s\circ r$ where $\circ: H\times H\to H$ denotes the multiplication of $R$ restricted to $H$.

Now to address your question: If for all $r\in H$ we have that $r\circ r=e$, then, $r=r^{-1}$ holds for all $r\in H$. Thus, given $r,s\in H$, $$r\circ s = r^{-1}\circ s^{-1}=(s\circ r)^{-1}=s\circ r,$$ where for the last equality we used that $s\circ r\in H$.

Answer 3

I assume you are using $e$ to denote the multiplicative identity of the ring $R$. I will follow custom and write $1$ rather than $e$. If $H$ is a subring of a ring $R$, then the additive identity $0$ is a member of $H$. As $0^2 = 0$, if $0^2 = 1$ we have $0 = 1$ which implies $x = 1x = 0x = 0$ for every $x \in R$. Thus your hypothesis implies that $R$ (and $H$) is the trivial ring with just one element $0 = 1$. The trivial ring is commutative.