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Part a) of Problem 2C.1 of Isaacs' Finite Group Theory reads

Show that every proper homomorphic image of an $N$-group is solvable.

What does "proper" mean here? Please note that I'm not asking for a solution to the problem, just a clarification of its statement, so I can work myself on it. (Incidentally, the definition of solvable group comes later in the book in Chapter 3, a rare oversight in Isaacs.)

Shaun
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Leandro Caniglia
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    I don't have the book to hand, but this most likely means a homomorphic image which isn't isomorphic to the group itself. – user1729 Jun 27 '23 at 15:45
  • A proper subset $N$ of $G$ is a subset $N$ which isn't the whole set $G$. A proper subgroup $N$ of $G$ is a subgroup with $N\neq G$. A nontrivial subgroup $N$ of $G$ is one with $N\neq 1$. – Dietrich Burde Jun 27 '23 at 16:10
  • @elemelons These are only necessarily equivalent for finite groups, as there are infinite groups which surject onto themselves with non-trivial kernel. So the phrase "proper homomorphic image" is surprisingly subtle! – user1729 Jun 27 '23 at 16:19
  • For a solution to the problem see https://math.stackexchange.com/a/4727047/269050. – Leandro Caniglia Jun 28 '23 at 15:52

1 Answers1

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A group $H$ is a homomorphic image of a group $G$ if there exists a surjective homomorphism $\phi:G\twoheadrightarrow H$. If $G$ is finite then $H$ is a proper homomorphic image if $G\not\cong H$; equivalently, if $\ker(\phi)$ is non-trivial.

As Isaac's book is about finite groups, this is the case here.

For infinite groups, these two conditions are not in general equivalent (see here) and so you should be careful using this term.

user1729
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