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While it's quite standard to infer the second isomorphism theorem from the first one, I've not seen the proof in reverse direction. Is it possible to infer the first isomorphism theorem from the second one? Thank you so much for your help!

I tried to let $S := G$ and $N := \ker(\varphi)$, but this leads to a trivial equality.

Here is first isomorphism theorem:

Let $\varphi: G \to H$ be a group homomorphism. Then $G/\ker(\varphi) \cong \operatorname{Im}(\varphi)$.

Here is second isomorphism theorem:

Let $G$ be a group, $S \le G$, and $N \trianglelefteq G$. Then $(S N) / N \cong S /(S \cap N)$.

Akira
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    The two statements are both true, so it doesn't really mean much to say that you can (or cannot) infer one from the other. Given any two theorems $A$ and $B$, there may or may not be "nice" ways of proving $A$ using $B$ as a lemma or vice versa. – Rob Arthan Jul 17 '20 at 21:53
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    In complete generality, the only options for $S$ are $1$, $N$ and $G$, and none of those works. So I don't think you can go the other direction, no. – David A. Craven Jul 17 '20 at 22:13
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    @MorganRodgers Could you please read the titles slowly? They are complerely different questions. If you agree with me, please remove your downvote. – Akira Jul 18 '20 at 05:51
  • To follow up on the comment of @RobArthan, here's part of the trouble. What does it mean to "infer the first isomorphism theorem from the second"? Perhaps it means that one accepts the second isomorphism theorem as an additional axiom of group theory, and in light of that one writes down a proof of the first isomorphism theorem. The trouble is, I could just ignore the light altogether and write down any proof of the first isomorphism theorem (in fact, I can even write a very short proof of the first isomorphism theorem, which ignores the light). – Lee Mosher Jul 18 '20 at 13:56
  • Hi @LeeMosher, Could you please transfer your comment into an answer so that I can accept it and remove it from list of unanswered questions? – Akira Jul 18 '20 at 14:08
  • I think the question is pretty clearly asking if there is a way to deduce the first from the second by a clever choice of $S$, $G$, and $N$, and without sneaking in a direct proof of the first. So I'm not sure these comments really answer the question. I'm inclined to agree with the spirit of David's remark, but one would need a more convincing argument that those are the only options. For example, given the assumptions of the first, one could apply the second to the group $G\times H$, with $S={(x,\varphi(x)):x\in G}$ and $N=\ker\varphi\times{1_H}$. Unfortunately, this doesn't work. – halrankard Jul 18 '20 at 14:11
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    Hi @halrankard That's exactly what I was thinking in writing down this question. – Akira Jul 18 '20 at 14:12

1 Answers1

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To follow up on the comment of @RobArthan, here's part of the trouble.

What does it mean to "infer the first isomorphism theorem from the second"?

Perhaps it means that one accepts the second isomorphism theorem as an additional axiom of group theory, and in light of that one writes down a proof of the first isomorphism theorem. The trouble is, I could just ignore the light altogether and write down any proof of the first isomorphism theorem (in fact, I can even write a very short proof of the first isomorphism theorem, which ignores the light).

Lee Mosher
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