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If I have two groups, $G$ and $H$ and two injective homomorphisms $\phi:G \to H$ and $\psi: H \to G$, then by the first isomorphism theorem applied to $\phi$, we have that $G \cong \mathrm{Im} (\phi)$, a subgroup of $H$. Similarly, $H$ is isomorphic to a subgroup of $G$. For finite groups, this guarantees that $G \cong H$ but does this hold for infinite groups?

Weird things can happen for infinite groups, e.g $n\mathbb Z \subsetneq \mathbb Z$ but $n \mathbb Z \cong \mathbb Z$. I'm wondering if this kind of thing stops an isomorphism from occurring.

I think this question generalises to rings, modules, fields and so on. Can this be answered for all of these structures?

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No. Any counterexample must be infinite, and one way to build such an example is to construct an analogue of the well-known injective homomorphism $S_n \hookrightarrow A_{n + 2}$ for the infinite analogues of those groups.

Consider the group $S_{\infty}$ of (finite) permutations of a set $\{1, 2, \ldots\}$ of countably many elements. Then, let $A_{\infty}$ denote the subgroup of $S_{\infty}$ comprised of even elements (products of even numbers of permutations), or equivalently, the subgroup generated by $3$-cycles. Then, inclusion $A_{\infty} \to S_{\infty}$ is a homomorphism.

On the other hand, let $S_{\infty}'$ denote the subgroup of $S_{\infty}$ of permutations of the set $\{3, 4, \ldots\}$, and consider the map $\Phi: S_{\infty} \to S_{\infty}'$ that maps a permutation $(a_1 \cdots a_n)$ to $((a_1 + 2) \cdots (a_n + 2))$ (this amounts to relabeling and so is manifestly an isomorphism). Then, we can define a map $\Psi: S_{\infty}' \to A_{\infty}$ by $$\Psi(\sigma) := \left\{ \begin{array}{cc} \sigma , & \sigma \text{ even} \\ \sigma (12), & \sigma \text{ odd} \end{array} \right. $$ and readily see that it is an injective homomorphism. So, $\Psi \circ \Phi$ is an injective homomorphism $S_{\infty} \to A_{\infty}$, and hence we have injective homomorphisms in both directions, but the two groups are not isomorphic.

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    $\begingroup$ Is there an easy proof that $A_\infty$ and $S_\infty$ are not isomorphic? $\endgroup$ – Christopher Apr 30 '15 at 18:47
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    $\begingroup$ @user73985 : consider the conjugacy class of transpositions in $S_\infty$. If you pick any two transpositions and get their product, you land into one of $3$ conjugacy classes (the trivial one, the $3$-cycles, and the double transpositions). However there is no such conjugacy class in $A_\infty$. For example if you pick the class of double transpositions, you can land into at least $6$ different conjugacy classes, and it only gets worse if you pick more complex ones. $\endgroup$ – mercio Apr 30 '15 at 19:21
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    $\begingroup$ The squares $\sigma^2$ of elements $\sigma \in S_{\infty}$ are all even, so they subgroup they generate is contained in $A_{\infty}$ and in particular proper. On the other hand, the squaring map sends each $3$-cycle $(i, j, k)$ to the respective $3$-cycle $(i, k, j)$ and in particular is a bijection on the set of $3$-cycles. Since $A_{\infty}$ is generated by $3$-cycles, the subgroup generated by the squares of the elements of $A_{\infty}$ is $A_{\infty}$ itself. $\endgroup$ – Travis Willse May 1 '15 at 3:38
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For groups this property - sometimes called the Cantor-Schröder-Bernstein property after the corresponding theorem for plain sets - is wrong. Let $G = F_2$ the free group on two generators $\{a,b\}$ and $H = F_3$ the free group on three generators $x,y,z$. Then there are monomorphisms $f \colon G \to H$, given by $f(a) = x$, $f(b) = y$ and $g \colon H \to G$ given by $g(x) = a^2$, $g(y) = b^2$, $g(z) = ab$, there is no isomorphism.

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I think this question generalises to rings, modules, fields and so on. Can this be answered for all of these structures?

The answer is no for rings and fields. It should be no for modules but I'm having trouble coming up with a snappy counterexample off the top of my head.

Rings: Let $k$ be a field, for simplicity. The ring $k[t^2, t^3]$ injects, by construction, into the ring $k[t]$. The ring $k[t]$ also injects into $k[t^2, t^3]$ via the map $t \mapsto t^2$. So there are injections

$$k[t] \hookrightarrow k[t^2, t^3] \hookrightarrow k[t]$$

but these two rings are not isomorphic because, for example, $k[t]$ is integrally closed and $k[t^2, t^3]$ is not.

Fields: The field $\mathbb{C}$ injects into the field $\mathbb{C}(t)$, and hence injects into its algebraic closure $\overline{ \mathbb{C}(t) }$. Now, fun theorem: (Edit: uncountable) algebraically closed fields of characteristic $0$ are completely determined by their cardinality, and $\overline{ \mathbb{C}(t) }$ has the same cardinality as $\mathbb{C}$. So they are isomorphic. So there are injections

$$\mathbb{C} \hookrightarrow \mathbb{C}(t) \hookrightarrow \mathbb{C}$$

but these two fields are not isomorphic because $\mathbb{C}$ is algebraically closed and $\mathbb{C}(t)$ is not.

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  • $\begingroup$ What about ordered fields? $\endgroup$ – celtschk Apr 30 '15 at 22:09
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    $\begingroup$ You mean uncountable algebraically closed fields. Look at the algebraic closure of $\Bbb Q(\pi)$ and that of $\Bbb Q$, both are countable and non-isomorphic. $\endgroup$ – Asaf Karagila May 1 '15 at 11:09
  • $\begingroup$ @celtschk: For ordered fields the analogue of algebraically closed is probably "real-closed fields" (in other words, the theory of $\Bbb R$), but that theory is not categorical in any cardinality. To see that, note that $\Bbb R$ and ${}^\ast\Bbb R$ (the hyperreals) are both of cardinality $2^{\aleph_0}$ but one is Archimedean and the other is not. $\endgroup$ – Asaf Karagila May 1 '15 at 12:27
  • $\begingroup$ @AsafKaragila: What does it mean for a theory to be categorical in a cardinality? $\endgroup$ – celtschk May 1 '15 at 13:22
  • $\begingroup$ @celtschk: Every two models of the same cardinality are isomorphic. For example every two vector spaces over a finite field are isomorphic if and only if they have the same cardinality. Over $\Bbb Q$ this is true for uncountable vector spaces. Similarly, every two countable dense linear orders without endpoints are isomorphic, so this theory is $\aleph_0$-categorical. For algebraically closed fields the theory is categorical in any uncountable cardinality, once you fix the characteristics. So every algebraically closed field extending $\Bbb Q$ of size $2^{\aleph_0}$ is isomorphic to $\Bbb C$. $\endgroup$ – Asaf Karagila May 1 '15 at 13:25
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A nonconstructive example: there are injective homomorphisms $$(\mathbb R,+)\to(\mathbb R\times\mathbb Z,+)\to(\mathbb R\times\mathbb R,+)\cong(\mathbb R,+)$$ but $(\mathbb R,+)\not\cong(\mathbb R\times\mathbb Z,+).$

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