To prove that if $f: G_1 \to G_2$ is an isomorphism and $N$ is a normal subgroup of $G_1$, then $f(N) = \{f(n) \mid n \in N\}$ is a normal subgroup of $G_2$, we proceed as follows:

1. Recall the definitions:

• $N \trianglelefteq G_1$ means $gNg^{-1} \subseteq N$ for all $g \in G_1$.
• $f: G_1 \to G_2$ is an isomorphism, so $f$ is bijective and satisfies $f(xy) = f(x)f(y)$ for all $x, y \in G_1$.
• To show $f(N) \trianglelefteq G_2$, we need to prove $g'f(N)(g')^{-1} \subseteq f(N)$ for all $g' \in G_2$, where $g' = f(g)$ for some $g \in G_1$.

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2. Image under $f$:

Define $f(N) = \{f(n) \mid n \in N\}$, which is a subset of $G_2$. Since $f$ is a bijection:

• $f(N)$ contains all elements of $G_2$ that correspond to elements of $N \subseteq G_1$.

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3. Normality condition in $G_2$:

Let $g' \in G_2$ and $f(n) \in f(N)$. We want to show: $g'f(n)(g')^{-1} \in f(N),$ where $g' = f(g)$ for some $g \in G_1$.

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4. Substitute using the properties of $f$:

Since $f$ is an isomorphism, it preserves group operations. Thus: $g'f(n)(g')^{-1} = f(g)f(n)f(g)^{-1}.$

Using the homomorphic property $f(xy) = f(x)f(y)$ and $f(x^{-1}) = (f(x))^{-1}$, the above becomes: $f(g)f(n)f(g)^{-1} = f(gng^{-1}),$ where $gng^{-1} \in N$ because $N$ is normal in $G_1$ ($N \trianglelefteq G_1$).

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5. Conclusion:

Since $gng^{-1} \in N$ and $f$ maps elements of $N$ to elements of $f(N)$, it follows that: $f(gng^{-1}) \in f(N).$

Thus: $g'f(n)(g')^{-1} \in f(N).$

Since this holds for any $g' \in G_2$ and any $f(n) \in f(N)$, we conclude: $f(N) \trianglelefteq G_2.$

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6. Final Statement:

If $N \trianglelefteq G_1$, then $f(N) \trianglelefteq G_2$, as required.

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Questions:

1. What does it mean for a subgroup to be normal in a group?
2. Why is it crucial that $f$ is an isomorphism in this proof?
3. How does the homomorphic property $f(xy) = f(x)f(y)$ ensure the preservation of group structure?
4. Can this result be generalized to automorphisms instead of isomorphisms? Why or why not?
5. How does the bijective nature of $f$ play a role in the proof?

Tip: Always ensure that when proving normality, the operations within the group align with the mapping properties (e.g., homomorphism or isomorphism).