Talk:Abstract Algebra/Group Theory/Normal subgroups and Quotient groups

Theorem 12
I don't think that theorem 12 holds. As a counterexample, I have found $$G = {0, 1}$$ with addition modulo 2 and the equivalence relation with $$0 \sim 1$$. Then $$G / \sim = [0]$$, and any Homomorphism $$\tilde \Phi : G / \sim \to H$$ must be the trivial homomorphism, because the only element in $$G / \sim$$ is zero. But the identity is not the trivial homomorphism and therefore can't be written in the form $$\tilde \Phi \circ \pi$$, because such a map would map any element to zero. So what am I missing? --Mathmensch (discuss • contribs) 16:07, 12 November 2013 (UTC)
 * Ah I see. I missed the assumption that $$a \sim b \Rightarrow \Phi(a) = \Phi(b)$$, which is not the case for the identity. --Mathmensch (discuss • contribs) 16:44, 12 November 2013 (UTC)

Lemma 16
Lemma 16 is patently false. Consider any group with at least three elements and the equivalence relations whose only two classes are the singleton of identity and its complement. Proof of one of the directions seems to only show that the quotient by a subgroup is a group iff the subgroup is normal. --87.99.27.160 (discuss) 02:26, 9 January 2015 (UTC)
 * Thanks for your comment! Will have a closer look on tuesday. --Mathmensch (discuss • contribs) 23:28, 16 January 2015 (UTC)
 * You are right. I will delete the lemma, or replace if necessary. --Mathmensch (discuss • contribs) 16:34, 19 January 2015 (UTC)
 * Replacement complete. --Mathmensch (discuss • contribs) 16:49, 19 January 2015 (UTC)

Lemma 13
I think that the equivalence relation in Lemma 13 should be such that $$a\sim b \,\Leftrightarrow ga\sim gb$$, not $$a\sim b \,\Leftrightarrow ga=gb$$, based on the proof, as well as Theorem 15. --212.251.76.205 (discuss) 07:15, 10 April 2017 (UTC)