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I am a complete beginner to Abstract Algebra. Today I asked my professor the following:

For group $G$ and $H$, let $f: G \to H$ be an homomorphism, and $N$ be a normal subgroup of $G$, if $\frac{G}{N} \cong Im(f)$, does that imply that $N \cong Ker(f)$.

After thinking for a few seconds, I think my professor said yes. I think by first isomorphism theorem $\frac{G}{Ker(f)} \cong Im(f)$. However, based on my search, in general for normal subgroup $H, N$$, \frac{G}{H} \cong \frac{G}{N}$ does not imply $H \cong N$, so I'm am not sure what would be the answer for what I asked.

Shaun
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wsz_fantasy
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    A useful perspective is that "normal subgroups" and "kernels of homomorphisms" are really exactly the same thing, and so are "quotients" and "images of homomorphisms" - this is what the first isomorphism theorem tells us. Mariano also assumes you understand this in his comment! In particular if you take any such counterexample where $G/H \cong G/N$ but $H \ncong N$, then let $f$ be the projection map $G \to G/H$. Can you argue that this gives a counterexample to your question? To be clear, I think your professor was wrong! (or perhaps they misunderstood what you were asking.) – Izaak van Dongen Sep 27 '23 at 02:20
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    My personal favourite counterexample is "Let $G_0$ be a nontrivial group. Let $G = G_0 \times G_0 \times G_0 \times \dotsb$. Let $N = 1$ be the trivial subgroup. Let $H = G_0 \times 1 \times 1 \times \dotsb$". These infinite products can be a good source of counterexamples, though if you need a finite counterexample Mariano's comment is more useful! Some sneaky cases of my example include $G = \mathcal P(\Bbb N)$, the powerset of $\Bbb N$ under symmetric differences with $H = \mathcal P({1})$, as well as $G = \Bbb Q^\times$, the rationals under multiplication, and $H = \langle 2 \rangle$. – Izaak van Dongen Sep 27 '23 at 02:31
  • Nor does $H \cong N$ imply $G/H \cong G/N$. For example, just let $G=\mathbb{Z}$ and $H$ and $N$ be any two distinct nonzero subgroups of $\mathbb{Z}$. – Geoffrey Trang Sep 27 '23 at 02:38
  • @IzaakvanDongen Thank you! Let me think about it. Can you elaborate on what you said by "normal groups" are the same as "kernels of homomorphism" though? – wsz_fantasy Sep 27 '23 at 02:45
  • I said normal subgroups are the same as kernels of homomorphisms. What I mean by this is that if $G$ is a group, then ${S \subseteq G \mid \text{$S$ is a normal subgroup of $G$}}$ and ${S \subseteq G \mid \text{there is some group $K$ and some hom $f: G \to K$ with $S = \ker f$}}$ are the same set! This is a corollary of the first isomorphism theorem, and I think one of the most important "morals" of the theorem. (well really the "quotients and images" version is more the corollary.. this version is a corollary of the fact that quotient groups exist) – Izaak van Dongen Sep 27 '23 at 02:49

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