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I'm new to the topic and I'm reading "Contemporary Abstract Algebra" by Gallian an this is how isomorphism is defined:

"An isomorphism $\phi$ from a group $G$ to a group $\overline{G}$ is a one-to-one mapping (or function) from $G$ onto $\overline{G}$ that preserves the group operation."

What is exactly meant by "preserves the group operation"? The one of $G$? $\overline{G}$?

Michael Albanese
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GniruT
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    As the answers show, both group operations on $G$ and $\overline G$ are involved. So it would be a little better to express this with the plural, "… preserves the group operations". Of course, writing it out in equation format as the answers do is completely unambiguous, but linguistic shorthands are common nonetheless. – Lee Mosher Nov 21 '15 at 14:50
  • Then this "preserves" means that $ \phi $ includes both operations on its own definition, right? – GniruT Nov 21 '15 at 15:29
  • I would word that a bit differently: both operations are included in the definition of the phrase ``$\phi$ is an isomorphism'' (not in the definition of $\phi$). – Lee Mosher Nov 21 '15 at 16:02
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    @LeeMosher Good point, but also "preserves the group operations" can be confusing for some readers: a group has only one operation, unlike for instance a ring, so one could be puzzled by the use of the plural. An alternative wording without these issues is "preserves the group structure". – Federico Poloni Nov 22 '15 at 07:13

4 Answers4

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If you have two elements $g_1, g_2 \in G$, there are two ways to produce an element in $\bar{G}$:

  1. use the binary operation $\ast$ of $G$ to produce the element $g_1\ast g_2$ of $G$, then apply the map $\phi$ to obtain $\phi(g_1\ast g_2) \in \bar{G}$.
  2. apply the map $\phi$ to both $g_1$ and $g_2$ to obtain $\phi(g_1), \phi(g_2) \in \bar{G}$, then use the binary operation $\cdot$ of $\bar{G}$ to produce the element $\phi(g_1)\cdot\phi(g_2) \in \bar{G}$.

The map $\phi$ is said to preserve the binary operation if it doesn't matter which of the above two processes we use. That is, $\phi$ satisfies $\phi(g_1\ast g_2) = \phi(g_1)\cdot\phi(g_2)$ for every $g_1, g_2 \in G$. In some sense, $\phi$ maps 'products' in $G$ to 'products' in $\bar{G}$.

Here is a more abstract way of viewing the above condition using the notion of transport of structure.

For convenience, let $b : G\times G \to G$ and $\bar{b} : \bar{G}\times\bar{G} \to \bar{G}$ denote the binary operations on $G$ and $\bar{G}$ respectively. As we have a bijection $\phi : G \to \bar{G}$ we can 'pullback' the binary operation $\bar{b}$ to a binary operation $\phi^*\bar{b}$ on $G$ as follows: for $g_1, g_2 \in G$, we define

$$\phi^*\bar{b}(g_1, g_2) := \phi^{-1}(\bar{b}(\phi(g_1), \phi(g_2))).$$

Now we have two binary operations on $G$, namely $b$ and $\phi^*\bar{b}$. The condition above is equivalent to the statement that $\phi^*\bar{b} = b$ (i.e. the two binary operations are the same).

In the special case where $\bar{G} = G$, we have $\bar{b} = b$ so the condition reads $\phi^*b = b$. That is, the group operation of $G$ is preserved by (the pullback via) $\phi$.

Michael Albanese
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  • But in which sense does $ \phi $ "preserve" the operations? Is it because it includes both in its own definition? – GniruT Nov 21 '15 at 15:26
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    @GniruT what Michael described in his answer is practically the definition of preserving an operation. Maybe you could clarify what about this still confuses you? – David Z Nov 21 '15 at 17:03
  • @DavidZ My confusion comes with the meaning of "preserve". For me it means that $\phi$ is associative, has neutral element and inverse i.e. it's still being an operation in a group. Otherwise, according to Michael "preserve the operation" has a mathematical meaning, which is what he just explained right? – GniruT Nov 21 '15 at 18:10
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    @GniruT: $\phi$ is not a binary operation. It cannot be associative or have a neutral element. It is a mapping (aka function) between two different sets. For a general map $f:G\to \bar{G}$, you don't necessarily have $f(g_1g_2)=f(g_1)f(g_2)$. But for a map that preserves the group operation, you do. – Meni Rosenfeld Nov 22 '15 at 00:08
  • Thanks for the explanation ^^ – GniruT Nov 24 '15 at 09:11
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Let $g\cdot h$ denote the multiplication in $G$ and $\overline g*\overline h$ be multiplication in $\overline G$. Then a homomorphism $f:G\to\overline G$ is a map such that

$$f(g\cdot h)=f(g)*f(h).$$

Tim Raczkowski
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It means that if $(G, \ast_{G},e_{G})$ and $(\overline{G}, \ast_{\overline{G}},e_{\overline{G}})$ are groups with their respective operation $\ast_{G}$ and $\ast_{\overline{G}}$. Then $\phi: G \to \overline{G}$ is an isomorphism if $\phi$ is bijective and for every $x,y \in G$:

$$\phi(x\ast_{G}y)=\phi(x)\ast_{\overline{G}}\phi(y).$$

The latter condition means that $\phi$ preserves the group operations.

Michael Albanese
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derthomas
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It means that $f(e_G)= e_{\overline G}$, and for all $x,y\in G, f(xy) = f(x)f(y)$, which together imply that $f(x^{-1}) = f(x)^{-1}$.

BrianO
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    I don't think the identity mapping is necessary. Indeed if $f(xy) = f(x)f(y)$ then we have $f(x) = f(e_G x) = f(e_G) f(x)$ and $f(x) = f(x e_G) = f(x) f(e_G)$ since this holds for all $x \in G$ and $f$ is one-to-one then it also holds for all $f(x) = y \in \overline{G}$ and hence $f(e_G)$ is the identity of $\overline{G}$. – Kayle of the Creeks Nov 21 '15 at 14:50
  • Why do you need that $f(x) = f(e_G x) = f(e_G) f(x)$ AND $f(x) = f(x e_G) = f(x) f(e_G)$ PLUS the hypothesis of bijectivity to have $f(e_{G})=e_{\overline{G}}$?

    Take f not bijective and only $$f(x) = f(e_G x) = f(e_G) f(x) \implies f(x)f(x)^{-1}=f(e_{G})f(x)f(x)^{-1} \implies e_{\overline{G}}=f(e_{G})e_{\overline{G}} \implies e_{\overline{G}}=f(e_{G})$$

    The inverse of $f(x)$ in $\overline{G}$ exists because $\overline{G}$ is a group, so where do we need bijectivity and the second equality $f(x) = f(x e_G) = f(x) f(e_G)$?

    Finally, how do we conclude that $f(x^{-1}) = f(x)^{-1}$?

     – derthomas Nov 21 '15 at 15:21
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    True, and bijectivity not required to infer that $f$ preserves the identity element. Just what the operation or operations plural of groups are, is of course relative to the axioms used. In the most redundant case, there are three: $e$, $*$, and $^{-1}$ (constants can be taken to be 0-ary operations). The axioms are such that when defining homomorphisms, it's redundant to specify that these are all preserved. For other kinds of structures, that isn't in general true. – BrianO Nov 21 '15 at 22:07
  • I see, and the inverse is preserved since

    $$f(e_{G})=f(xx^{-1})=f(x)f(x^{-1})$$

    now multiply left both sides by $f(x)^{-1}$ and you get $f(x)^{-1}=f(x^{-1})$.

    Is this the right terminology "multiply left"?

     – derthomas Nov 22 '15 at 11:21
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    Yes, that's correct, and it's the correct terminology.. I would have said "right" but didn't want to be confusing - left/right & so on :) – BrianO Nov 22 '15 at 11:46
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    I was also going to add **pun intended** at the end of my last sentence ;) – derthomas Nov 22 '15 at 12:35
  • I for one would have laughed. And given a shorter answer ;/ – BrianO Nov 22 '15 at 21:00