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p. 128, 129. Theorem 13.12. Let $h$ be a homomorphism of groups $G \to G'$.
III. If $S \le G$, then $h[S] \le \color{red}{G'}$.
IV. If $S' \le G'$, then $h^{-1}[S'] \le G$.

p. 149. Theorem 15.16. Let $h$ be a homomorphism of groups $G \to G'$.
III. If $N \unlhd G$, then $h[N] \unlhd \color{red}{h[G]}$.
IV. If $N' \unlhd \color{red}{h[G]}$, then $h^{-1}[N'] \unlhd G$.

Note that $h[N]$ may not be normal in G', even though $N \unlhd G$. For example, define $h: \mathbb{Z}_2 \to S_3$ by means of $h(0) = id$ and $h(1) = (1)(2,3)$. h is a homomorphism. Any group is a normal subgroup of itself, but $\{id , h(1)\}$ is not a normal subgroup of $S_3$.

(1.) What's the intuition behind the first difference in red? Why $h[G]$ instead of just $G'$?

(2.) What's the intuition behind the second difference in red?

(3.) How do you envisage and envision Fraleigh's counterexample for $h[N] \not \unlhd G'$?

(4.) To prove $\{id , h(1)\} \not \unlhd S_3,$ I can use the Normal Subgroup Test.
Hence I need to find exactly one $g \in S_3$ so that $g\{id , h(1)\}g^{-1} \not \in S_3$.
How do you find $g$ craftily, without trial and error? My Cayley diagram by dint of Group Explorer:

enter image description here

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If $S\leq G$, then $h[S] \leq G'$: This is because the map is a homomorphims.ie It preseves the fact that a subgroup is a group.

If $N⊴G$, then $h[N]⊴ h[G]$: Normality of a subgroup depends on the group containing it. The group structure of $G$ is preserved by $h$ and so it preserves its subgroup structures as well. The third case is similar.

Lastly, to show the group is not normal, recall that conjugacy classes in $S_3$ can be seen from cycle type. In particular, if you conjugate by an element with different cycle type e.g $(123)$, you'll leave the subgroup.

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  • $\begingroup$ The first part doesn't make sense. If $h\colon G\to G'$ then $h[S]$ is by definition contained in $G'$, not $G$. $\endgroup$ – Hagen von Eitzen Feb 21 '14 at 19:01
  • $\begingroup$ that must be a typo $\endgroup$ – Hesky Cee Feb 21 '14 at 19:04
  • $\begingroup$ @IhechukwuChinyere: your comment seems to answer the question. You answer seems to miss the point. $\endgroup$ – robjohn Feb 24 '14 at 17:55

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