MHB What Determines Group Homomorphisms Between Rational Numbers?

[Fermat1]

1)Describe all group homomorphisms from (Q,+) to (Q,+) where Q is the set of rationals.

Googling this, I came across the fact that for integer $n$, $f(n)=f(0+n)=n+f(0)$

But I don't understand the third step. To my mind $f(0+n)=f(0)+f(n)=f(n)$

2)Is there a surjective homomorphism from $Q$ to $Z_{2}$?

Is there a subgroup of $Q$ of index 2?

Thanks

 

[Euge]

1)Describe all group homomorphisms from (Q,+) to (Q,+) where Q is the set of rationals.

Googling this, I came across the fact that for integer $n$, $f(n)=f(0+n)=n+f(0)$

But I don't understand the third step. To my mind $f(0+n)=f(0)+f(n)=f(n)$

2)Is there a surjective homomorphism from $Q$ to $Z_{2}$?

Is there a subgroup of $Q$ of index 2?

Thanks

Hi Fermat,

You're right to find an error in $f(n) = f(0+n) = n+f(0)$. It turns out that the group homomorphisms are of the form $f_a : x \to ax$ for all $a \in \Bbb Q$. To show this, set $a = f(1)$. Note that $a$ can be any rational number. For all $n \in \Bbb N$, $f(n) = nf(1) = na$. Since $f(0) = 0$, $f(-x) = -f(x)$ for all $x\in \Bbb Q$. Therefore $f(-n) = -f(n) = -na$ for all $n\in \Bbb N$. Thus $f(n) = na$ for all $n\in \Bbb Z$. Given $m\in \Bbb N$, $f(1) = f(m \cdot \tfrac{1}{m}) = mf(\tfrac{1}{m})$. Thus $f(\tfrac{1}{m}) = \frac{1}{m}f(1)$ whenever $m\in \Bbb N$. Consequently, for all $n\in \Bbb Z$ and $m\in \Bbb N$, $f(\tfrac{n}{m}) = nf(\tfrac{1}{m}) = \frac{n}{m}a$. Hence, $f(x) = ax$ for all $x\in \Bbb Q$.

The answer to the second question is no. Having a surjective homomorphism $f$ of $\Bbb Q$ onto $\Bbb Z_2$ implies (by the first isomorphism theorem) that $\Bbb Q/N$ has order 2, where $N = \text{ker}f$. Since $2\overline{x} = \overline{0}$ in $\Bbb Q/N$, $2x \in N$ for all $x \in \Bbb Q$. Thus $2\Bbb Q \subseteq N$. Since $2\Bbb Q = \Bbb Q$, $\Bbb Q \subseteq N$. Therefore $\Bbb Q = N$ and $\Bbb Q/N$ has one element, a contradiction. Note that this also shows that $\Bbb Q$ does not have a subgroup of index 2.

 

[Fermat1]

Hi Fermat,

You're right to find an error in $f(n) = f(0+n) = n+f(0)$. It turns out that the group homomorphisms are of the form $f_a : x \to ax$ for all $a \in \Bbb Q$. To show this, set $a = f(1)$. Note that $a$ can be any rational number. For all $n \in \Bbb N$, $f(n) = nf(1) = na$. Since $f(0) = 0$, $f(-x) = -f(x)$ for all $x\in \Bbb Q$. Therefore $f(-n) = -f(n) = -na$ for all $n\in \Bbb N$. Thus $f(n) = na$ for all $n\in \Bbb Z$. Given $m\in \Bbb N$, $f(1) = f(m \cdot \tfrac{1}{m}) = mf(\tfrac{1}{m})$. Thus $f(\tfrac{1}{m}) = \frac{1}{m}f(1)$ whenever $m\in \Bbb N$. Consequently, for all $n\in \Bbb Z$ and $m\in \Bbb N$, $f(\tfrac{n}{m}) = nf(\tfrac{1}{m}) = \frac{n}{m}a$. Hence, $f(x) = ax$ for all $x\in \Bbb Q$.

The answer to the second question is no. Having a surjective homomorphism $f$ of $\Bbb Q$ onto $\Bbb Z_2$ implies (by the first isomorphism theorem) that $\Bbb Q/N$ has order 2, where $N = \text{ker}f$. Since $2\overline{x} = \overline{0}$ in $\Bbb Q/N$, $2x \in N$ for all $x \in \Bbb Q$. Thus $2\Bbb Q \subseteq N$. Since $2\Bbb Q = \Bbb Q$, $\Bbb Q \subseteq N$. Therefore $\Bbb Q = N$ and $\Bbb Q/N$ has one element, a contradiction. Note that this also shows that $\Bbb Q$ does not have a subgroup of index 2.

Thanks but I don't understand $f(n.1)=nf(1)$ Are you viewing it is a vector space over Z? If so you couldn't you view it as a vector space over Q?

 

[Fermat1]

solved it. Thanks

 

[mathbalarka]

Thanks but I don't understand $f(n.1)=nf(1)$ Are you viewing it is a vector space over Z? If so you couldn't you view it as a vector space over Q?

Follows from the addition-preserving property of homomorphisms

$$f(n \cdot 1) = f(1 + 1 + 1 + \cdots + 1) = f(1) + f(1) + f(1) + \cdot + f(1) = nf(1)$$

 

[Deveno]

Thanks but I don't understand $f(n.1)=nf(1)$ Are you viewing it is a vector space over Z? If so you couldn't you view it as a vector space over Q?

Sort of. The correct term is "$\Bbb Z$-module". Modules are "like" vector spaces, but the "scalars" are elements of a RING, not a field.

Any abelian group $G$ comes with a natural action ("scalar multiplication") of the integers on it:

For $k \in \Bbb Z$ and $g \in G$, we set:

$k\cdot g = g^k$.

If the "group operation" is written as addition, then $g^k$ is usually written $kg$, meaning:

$g + g +\cdots + g$ ($k$ times).

Here's the thing: any group homomorphism $f: \Bbb Q \to \Bbb Q$, is completely determined by $f(1)$. Euge's post shows why this is so. Using the $\Bbb Z$-module structure allows us to "sneak in multiplication".