Prove that $(\mathbb{Q}, +)$ is not isomorphic to $(\mathbb{Q}^n, +)$

Question

There are a number of solutions that I've seen to this question but someone showed me a solution where the argument was as follows:

If $h,k \in \mathbb{Q}$ such that $h \neq 0$ and $k \neq 0$, then $\langle h\rangle \cap \langle k\rangle \neq \{0\}$.

(This is an easy enough property to prove.)

But in $\mathbb{Q}^n$, where $\mathbb{Z}^n$ is a subgroup, the elements $a = (1,0,0,...,0)$ and $b=(0,0,...,0,1)$ are nontrivial but $\langle a\rangle \cap \langle b\rangle = (0,0,...,0)$ is trivial.

Therefore, $\mathbb{Q}$ and $\mathbb{Q}^n$ cannot be isomorphic as additive groups.

I do not understand why this proves that the two groups are not isomorphic? Can someone please explain this to me please?

Thank you kindly.

Best Regards.

Answer 1

"Transport of structure" is an extremely important thing to understand in math. Unfortunately the current Wikipedia article on it looks intimidating. An isomorphism can be used to turn properties of, or statements about, one algebraic structure into the same properties of, or statements about, another algebraic structure.

Consider properties of or statements about the ring of integers. Is multiplication of integer commutative only because we represent them with numerals like $1$, $2$, $3$? Would multiplication lose the commutative property if we instead represented integers with English words like one, two, three? Consider the statement that there is no integer which, when added to itself, yields the multiplicative identity. Could there suddenly exist such an integer if we represented them in Spanish like uno, dos, tres? No, the labels do not affect the truth or falsity of statements directly about the algebraic structure.

(Of course, it can affect non-mathematical statements, like "one is represented with three letters," or external statements like "{one, two, three} is a subset of $\{1,2,3,4,\cdots\}$." Statements or properties which can be expressed in the language of algebra and without reference to external objects are the focus.)

How the elements of the algebraic structure are named or referenced or labelled doesn't actually affect the algebra. So it is no surprise that relabelling the elements $-$ i.e. an isomorphism $-$ does not affect the truth or falsity of statements about the algebraic structure. If $\Bbb Q^n$ has cyclic subgroups which intersect trivially, and there were an isomorphism $\Bbb Q^n\cong\Bbb Q$, then that isomorphism would turn those trivially-intersecting cyclic subgroups of $\Bbb Q^n$ into trivially-intersecting cyclic subgrousp of $\Bbb Q$, which is impossible.

In more detail: if $f:G\to H$ is a homomorphism of groups, then applying $f$ to a cyclic subgroup $\langle g\rangle$ of $G$ yields a cyclic subgroup $\langle f(g)\rangle$ of $H$. In other words, $f\big(\langle g\rangle\big)=\langle f(g)\rangle$. Similarly, if $f:X\to Y$ is any bijection between sets, then $f(A\cap B)=f(A)\cap f(B)$. If you draw a Venn diagram, relabelling things that are in the diagram doesn't change where they are in the diagram (assuming distinct things get distinct labels, of course - as in a bijection). So in your case, if $\langle e_1\rangle\cap\langle e_2\rangle=\{\vec{0}\}$ in $\Bbb Q^n$ and there is an isomorphsim $f:\Bbb Q^n\to\Bbb Q$, we can apply $f$ to both sides and get $\langle f(e_1)\rangle\cap\langle f(e_2)\rangle=\{0\}$, with $f(e_1),f(e_2)\ne0$, which is a contradiction since we know (nontrivial) cyclic subgroups of $\Bbb Q$ do not intersect trivially.

Answer 2

Suppose there were an isomorphism $f : {\bf Q}^n \to {\bf Q}$.

Then $\langle f(a)\rangle $ and $\langle f(b)\rangle $ are subgroups of ${\bf Q}$, and their intersection is a subgroup of ${\bf Q}$ as well. We know that this intersection must contain a nonzero element, call it $c$. Then $f^{-1}(c)$ must be a nonzero element of ${\bf Q}^n$. But in fact $\langle a\rangle $ and $\langle b\rangle $ have trivial intersection, so we must have $c = 0$, a contradiction.

In effect, an isomorphism induces another isomorphism on the lattice of subgroups of both groups.

Answer 3

Explain the property in words: in $\mathbf Q$, all pairs of nonzero cyclic subgroups have a nontrivial intersection (they contain a common nonzero integer), but in $\mathbf Q^n$ for $n \geq 2$ there is a pair of nonzero cyclic subgroups with a trivial intersection, and that's the example you mention.

A similar method can distinguish between $\mathbf Q^m$ and $\mathbf Q^n$ for different positive integers $m$ and $n$ because the group structure of $\mathbf Q^n$ allows us to reconstruct what $n$ must be. Here's how to do that. Consider the following property $P_k$: there is a set of $k$ nonzero cyclic subgroups $H_1, \ldots, H_k$ such that each $H_i$ has a nonzero intersection with $\sum_{j \not= k} H_j$. Property $P_k$ is true in $\mathbf Q^n$ for $k \leq n$ while property $P_{k}$ is false in $\mathbf Q^n$ for all $k > n$. This is due to the fact that $\mathbf Q^n$, as a $\mathbf Q$-vector space, has linearly independent subsets of each size less than or equal to $n$ and no linearly independent subset of size greater than $n$. (I defined $P_k$ as a purely group-theoretic property, but it's really a property about linear dependence and independence in $\mathbf Q$-vector spaces: I've described the condition that $\mathbf Q^n$ has dimension $n$ using only group theory.) So we can characterize the $n$ in $\mathbf Q^n$ by saying it's the largest positive integer $k$ such that $P_k$ is true in $\mathbf Q^n$.

Answer 4

To prove that two things are isomorphic, all you need to do is to find an isomorphism. But to prove that two things are not isomorphic, you need to find some property or invariant that is preserved by isomorphism but which is different for your two things.

For example, if you want to show two sets aren't isomorphic, you can look at their cardinality. If you want to show that two vector spaces aren't isomorphic, you can look at their dimension. To tell two topological spaces aren't isomorphic, you can look at their fundamental groups or homology groups or cohomology rings or other such invariants.

This isn't the only way, as sometimes you can get a proof by contradiction that doesn't require an invariant or a property, for example things related to Cantor's diagonal argument, but invariants and properties are used more commonly.

In this particular case, the property we want to use is "Every two non-trivial subgroups have non-trivial intersection." This is a property preserved by isomorphism, and which $\mathbb Q$ satisfies but $\mathbb Q^n$ does not.