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Given two isomorphic groups G and H, is it possible that two or more functions define their isomorphism? Also, is it possible that another group say, L is isomorphic to G but not to H?

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Indeed a single group $G$ may have more than one automorphism. – hardmath Jul 22 '14 at 4:34
Recall that isomorphism is a equivalence relation. Thus if $G \cong H$ and $H \cong L$ then we have $G \cong L$ by transitivity. A group isomorphism is not necessarily unique. – John Smith Jul 22 '14 at 17:13
up vote 8 down vote accepted

Take $G=\Bbb Z/3\Bbb Z$ and $H=\{1,x,x^2\}$ where $x^3=1$. Then there are two isomorphisms here, you can send $x\mapsto [1]$ or $x\mapsto [2]$, both are isomorphisms. However, you cannot have $G\cong H, H\cong L, G\not\cong L$, being isomorphic is an equivalence relation, so it's transitive. In particular, if you have isomorphisms:

$$\begin{cases}\phi : \;G\to H \\ \psi:\; H\to L\end{cases}$$

then $\psi\circ \phi: G\to L$ is also an isomorphism.

As a generalization, if $\varphi\in\text{Aut}(G)$, the automorphism group of $G$, then all isomorphisms $f: G\to H$ give rise others via $f'=\varphi\circ f$, and it is easily shown that all isomorphisms between the two are of this form for some choice of $\varphi$.

As Micah notes in the comments, if you're looking for a trivial automorphism group, you're going to be disappointed unless you're into finite, simple groups of order 2. I'll plagiarize this source and just quote

(1) The group must be Abelian: Otherwise, it would have nontrivial inner automorphisms.

(2) The group must have exponent two: Otherwise, the map sending an element to its inverse, would be a nontrivial automorphism. In particular, it is an elementary Abelian group, or is a vector space.

(3) The group must be one-dimensional as a vector space, otherwise it will have other automorphisms (the general linear group on a vector space of dimension greater than one, is nontrivial)

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Is there a way to check if isomorphism is unique? – jevie Jul 22 '14 at 4:42
Yes, by my second comment, an isomorphism is unique, iff the automorphism group of $G$ is trivial. – Adam Hughes Jul 22 '14 at 4:43
...which almost never happens. @jevie – Micah Jul 22 '14 at 4:45
haha, yes, I should have included that in the answer since it's so simple. Thanks for mentioning it @Micah – Adam Hughes Jul 22 '14 at 4:48
This was the faster answer, and with the inclusion of that last proof, I'd probably have accepted it over mine myself. – JHance Jul 22 '14 at 5:19

To the first, yes, to the second, no. I'll address the second question first. An isomorphism can be succinctly characterized as a homomorphism $\phi: G \to H$ which is invertible (it is a nice property of homomorphisms and algebraic structure maps in general that the set theoretic inverse of a bijective homorphism is still a homomorphism). If you accept that the composition of two homomorphisms is still a homomorphism, then we we can invert the composition $\psi\phi: G \to L$ of two isomorphisms $\phi: G \to L$, and $\psi: H \to L$ by just taking the reverse composition $\phi^{-1}\psi^{-1}$. In particular, isomorphism is an equivalence relation.

To the second, the uniqueness of isomorphisms $G \to H$ is basically classified by the automorphisms $$\text{Aut}(G) = \{\psi:G \to G| \psi \text{ an isomorphism}\}$$

It is worthwhile to check that the above set is a group under composition of functions. In particular, given a particular isomorphism $\phi: G \to H$, we can define a new isomorphism $G \to H$ by simply composing with an element of the automorphism group $\phi \circ \psi$. In fact, this is a bijection, though it is not in any way canonical. There are exactly as many isomorphism $G \to H$ as there are automorphisms of $H$. (This makes some sense if we think of $G,H$ as being essentially the same group up to a relabelling)

A final note, which is the death knell of the uniqueness of isomorphisms except in two very small cases: There are no groups with non-trivial automorphism group except the trivial group $\{e\}$ and $\mathbb{Z}/2\mathbb{Z}$.

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