Is $\{0,1\}$ a subgroup of $\mathbb{R}$ under multiplication? I am playing around with subgroups and I was wondering if the group $\{0,1\}$ is a subgroup of $\mathbb{R}$ with respect to multiplication. It seems to  fit the criteria but the inverse property is making me worry because $0$ is its own inverse. I vaguely remember a proof that if an element was its own inverse than it must be the identity element  but $1$ would need to be the identity here. Thanks in advance! 
 A: Let's begin from the fact that $\mathbb{R}$ is not a group with respect to multiplication because $0$ has no inverse at all. So of course it can't have any subgroups. Now, if you look at $\mathbb{R}\setminus\{0\}$ then it is a group with respect to multiplication. And no, it is not true that if an element is its own inverse then it is the identity element. For example in the group I mentioned $(-1)$ is its own inverse. 
A: To reiterate Mark's good answer, $\{0,1\}$ is not a subgroup of $\mathbb{R}$ for two reasons:

*

*$\mathbb{R}$ is not a group under multiplication. In order for $A$ to be a subgroup of $B$, $B$ has to be a group.
($\mathbb{R}$ is only a group under addition. Also, $\mathbb{R} \setminus \{0\}$ is a group under multiplication. But $\mathbb{R}$ under multiplication isn't.)


*$\{0,1\}$ is not a group under multiplication itself.

It seems to  fit the criteria but the inverse property is making me worry because 0 is its own inverse. I vaguely remember a proof that if an element was its own inverse than it must be the identity element  but 1 would need to be the identity here.

Partly correct. $0$ isn't its own inverse -- $1$ is the only identity element in $\{0,1\}$ with addition, and $0$ isn't an identity, so $0$ wouldn't be called its own inverse.
But what is true -- and seems close to the reasoning you are getting at -- is that if $a \cdot a = a$ in a group, then $a$ is the identity. This is because you can multiply on both sides by $a^{-1}$ and then you get $a = e$ (where $e$ denotes the identity).
Here, $0 \cdot 0 = 0$, so $0$ would have to be the identity. But we know that doesn't work, since $1$ is already the identity, and $0 \cdot 1 = 0$, not $1$.
A: Hint: The Cayley table of $\{0, 1\}$ under multiplication is
$$\begin{array}{c| c c}
\times & 0 & 1\\
\hline
0 & 0 & 0 \\
1 & 0 & 1.
\end{array}$$
What do you know about Latin squares?
