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I was aware that homomorphism/isomorphism is order-preserving (i.e. $\phi: G \to H$ is an isomorphism $\implies$ ord(g) = ord($\phi(g))$, $\forall g \in G$), but I was not sure whether the converse holds (for finite groups):


Let G, H be groups.

If ord(G) = ord(H), and $\forall$ g $\in$ G, $\exists$ (or $\exists!$) h $\in$ H s.t. ord(g) = ord(h).

(And vise versa, $\forall$ h $\in$ H, $\exists$ (or $\exists!$) g $\in$ G s.t. ord(g) = ord(h).)

Does this imply that G $\cong$ H please?

If not, what will be a counterexample please?


More rigorously, suppose $f: H \to G$ is a bijection s.t. ord(f(x)) = ord(x), $\forall$ x $\in$ G.

Does this imply that f an isomorphism please?


I think it doesn't hold for infinite groups, but for finite groups, I was struggling to either find a direct proof or a counterexample.

I have tried constructing counterexamples by taking the direct product of some non-abelian & non-cyclic groups of lower order, but didn't quite work out.

An example which I considered is $Z_3 \times Z_3 \times Z_3$ and G = <x, y, z> modulo $x^3=y^3=z^3$; yz = zyx; xy = yx; xz= zx

Many thanks in advance!

Newton
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  • Well at the very least it doesn't work (for isomorphisms) when you don't specify uniqueness of existence. For example $G = (\mathbb{Z}/2\mathbb{Z})^n$ and $H = \mathbb{Z}/2\mathbb{Z}$ – Vercingetorix Oct 15 '22 at 00:47

2 Answers2

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This is quite false even for finite groups, see this MO question. A nice counterexample is given by the Heisenberg group $H_3(\mathbb{F}_3)$, which is a group of order $27$ where all non-identity elements have order $3$, so has the same order profile as $C_3^3$ (which is stronger than your first condition; we remember multiplicities, or equivalently there is a bijection which preserves orders, which is your second condition), but the two are non-isomorphic because the Heisenberg group is non-abelian. The smallest counterexamples have order $16$, and a general counting argument shows that the number of $p$-groups grows much too fast for any statement like this to be true.

Qiaochu Yuan
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  • This is a nice counteexample! Thank you :) – Newton Oct 15 '22 at 15:54
  • Could you elaborate on the ‘counting argument’ please? – Newton Oct 17 '22 at 19:35
  • @Newton: you can see the MO link for details. Basically the number of groups of order $p^n$ for $p$ a prime is known to grow like $p^{ \frac{2}{27} n^3}$ (with a large error term in the exponent) and this is much faster growth than the number of possible order profiles of such a group. – Qiaochu Yuan Oct 18 '22 at 00:59
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A simple counterexample is $\mathbb{Z}$ and $\mathbb{Z}\oplus\mathbb{Z}$. If you are looking for a counterexample in finite case, then there are more sophisticated examples, see this post: Does the order, lattice of subgroups, and lattice of factor groups, uniquely determine a group up to isomorphism?

Denote by $sub(G)$ a collection of all proper subgroups of $G$. The post above shows that there are two finite groups $G$ and $H$ such that there is a bijection $f:sub(G)\to sub(H)$ such that $X\simeq f(X)$, yet $G$ and $H$ are not isomorphic. The post above gives example of even stronger condition, that additionaly factor groups are isomorphic.

Yes, groups are weird.

freakish
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