Define group structure on $\mathbb{R \setminus Q}$

Question

Is it possible to define a group structure on $\mathbb{R \setminus Q}$ ? If yes how? $\mathbb{R}$ & $\mathbb{R \setminus Q}$ have the same cardinality but if I pull the group structure back to $\mathbb{R \setminus Q}$ I should have a group homomorphism first to make $\mathbb{R \setminus Q}$ a group!

Answer 1

If you mean "$\mathbb{R}/\mathbb{Q}$," then the answer is yes, since $(\mathbb{Q}, +)$ is a normal subgroup of $(\mathbb{R}, +)$.

If you really mean "$\mathbb{R}\setminus\mathbb{Q}$," though - that is, the set of irrationals - then the answer is still yes, it's just more messy. For example

• We can define a bijection between $\mathbb{R}$ and the irrationals, and pull the usual additive group structure on $\mathbb{R}$ across via this bijection. (Note that we could do a similar thing for $\mathbb{R}/\mathbb{Q}$ above, but it's less natural than the quotient group in that case.)

• We can also define (using e.g. continued fractions) a bijection between the irrationals and the set of infinite sequences of integers (basically, Baire space). The latter is a group via componentwise addition, and we can pull this group structure over to the irrationals. This has the advantage of being topologically nice: the irrationals are actually homeomorphic to Baire space, so in a sense we're not really doing too much damage (unlike the example above).

Other than these (+ similar), I don't know of any off the top of my head - and indeed, we can prove that there can't be really nice ones in the sense that there are none which "come from" groups on $\mathbb{R}$:

Suppose $*$ is a group operation on $\mathbb{R}$. Then $(\mathbb{R}\setminus\mathbb{Q}, *)$ is not a subgroup - in particular, it is not closed under $*$.

Proof. For $x\in\mathbb{R}$ let $\hat{x}$ be the unique real such that $\hat{x}*x=0$ (this will always exist since $*$ is a group operation). If the irrationals formed a subgroup, we would have to have $\hat{x}\in\mathbb{Q}$ for every irrational $x$. But then since there are uncountably many irrationals, there would be two distinct irrationals $x, y$ with $\hat{x}=\hat{y}$ - and this can't happen if $(\mathbb{R}, *)$ is a group.

Of course, there's nothing special about $\mathbb{R}$ here - the point is that if $A$ is a subgroup of $B$, then the cardinality of $B\setminus A$ must be at least the cardinality of $A$.

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Looking at your question, it seems you are confused about pulling group structures back along maps. You write

if I pull the group structure back to R\Q i should have a group homomorphism first to make R\Q a group

but this is getting the order of things wrong: the group structure on the irrationals comes from the choice of bijection, and so the bijection is automatically a homomorphism. Abstractly, what we have is:

Suppose $(X, *)$ is a group and $f: X\rightarrow Y$ is a bijection. Let $\cdot$ be the operation on $Y$ given by $$a\cdot b=c\iff f^{-1}(a)*f^{-1}(b)=f^{-1}(c).$$ Then $(Y, \cdot)$ is a group, and $f$ is a group homomorphism.

This is a good exercise, and if you do it it should clear things up a lot.