This is Exercise 5 on page 60 of Analysis I by Amann and Escher.
Exercise:
Let $\varphi \colon G \to G'$ be a homomorphism and $N'$ a normal subgroup of $G'$. Show that $\varphi^{-1}(N')$ is a normal subgroup of $G$.
Discussion:
I found drhab's answer here and he shows that normality can be proven in about two lines. My question is whether this can be proven by showing equality of left and right cosets. My text (which is admittedly focused on analysis) defines a normal subgroup as one for which left and right cosets are equal. It does not discuss conjugation (is that the right word?) at all.
I found trying to prove that $g \odot \varphi^{-1}(N') = \varphi^{-1}(N') \odot g$ for $g \in G$ to be quite difficult. I had no idea how to use the normality of $N'$, which I assumed was necessary.
Does anyone know how to prove this using cosets? I appreciate any help.
Best Answer
Here's an approach that tries hard not to use anything equivalent to conjugates. (For the record, conjugates are definitely the way to go.)
The main idea is $$g \odot \phi^{-1}(N') = \phi^{-1}(\phi(g) \odot N') = \phi^{-1}(N' \odot \phi(g)) = \phi^{-1}(N') \odot g.$$ The first and last equalities are sketchy and need justification. I'll do the first one since the second is the same.
For $x \in G$, we have $$\begin{align*}x \in \phi^{-1}(\phi(g) \odot N') &\Leftrightarrow \phi(x) \in \phi(g) \odot N' \\ &\Leftrightarrow \phi(g)^{-1} \odot \phi(x) \in N' \\ &\Leftrightarrow \phi(g^{-1} \odot x) \in N' \\ &\Leftrightarrow g^{-1} \odot x \in \phi^{-1}(N') \\ &\Leftrightarrow x \in g \odot \phi^{-1}(N'). \end{align*}$$