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I just re-read Herstein's popular text Topics in Algebra, where he says that an isomorphsim between two groups is defined as a homomorphism between those two groups with the added property of injectivity.

However, the very next definition shows that two groups are said to be isomorphic iff there exists an isomorphism from one of the groups onto the other. And, many standard perception and texts say that isomorphism is always bijective. So, what is this thing all about? Is injectivity sufficient for most properties of isomorphism, like, the isomorphism/homomorphism theorems? Any hints? Thanks beforehand.

Shaun
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vidyarthi
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2 Answers2

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Herstein's definition of a group isomorphism is no longer the standard definition. He defines an isomorphism to be a homomorphism that is one-one, and his meaning of one-one is what we now call injective. That is not the same as the current standard definition of an isomorphism, which is a bijective homomorphism.

Derek Holt
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    I should add that Herstein's book is still an excellent course on algebra, particularly suitable for able students, but it has to be read with caution, because some of his terminology is no longer standard. Also he never explicitly introduces the topic of groups on acting on sets, which plays a central role in all current courses. – Derek Holt Jan 08 '24 at 11:11
  • I think the root of this issue is that Herstein is using the older definition of a function, where we identify a function with its graph (such a definition remains popular in some branches of mathematics, particularly set theory and mathematical logic). If you define a function $f$ as a set of ordered pairs such that if $(x,y)\in f$ and $(x,y')\in f$, then $y=y'$, then there is no such thing as the "codomain" of a function. – Joe Jan 09 '24 at 01:36
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    For example, the maps $f:\mathbb R\to\mathbb R$ given by $f(x)=x$ and $g:\mathbb R\to\mathbb C$ given by $g(x)=x$ are the same set, and so it makes no sense to assert that $f$ is surjective while $g$ is not. If you define a function in this way, then Herstein's definition is actually completely correct. But in algebra, the "set of ordered pairs definition" of a function is not as useful as the modern one. – Joe Jan 09 '24 at 01:38
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    There has been some discussion of this on Mathematics Stack Exchange (see When do two functions become equal?). – Joe Jan 09 '24 at 01:40
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Herstein is introducing an isomorphism between $R$ and $R'$ as an injective homomorphism which does not imply, as we see by his next definition, that $R$ and $R'$ are isomorphic.

This is a slightly odd way to introduce things and was probably considered standard at the time. I have only ever seen isomorphism defined as a bijective homomorphism and of course, two groups defined as isomorphic if there is a bijective homomorphism between them. This seems to be the modern way of introducing the topic.

Edit: I have since checked 'Theory of Groups' by Hall which was released in 1959, 5 years earlier than 'Topics in Algebra' and it too introduces an isomorphism as an injective homomorphism.

rangen1
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