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This question has already been answered here: Prove that $gNg^{-1} \subseteq N$ iff $gNg^{-1} = N$

I was wondering whether the following proof is also acceptable and if not what is wrong with it. Suppose that $gNg^-1 \subset N$ but there exist $n_1$ s.t $n_1 \in N$ but $n_1 \notin gNg^-1$. Since $gNg^-1 = \{gng^-1 | n \in N\}$ we know that $\exists \ a \in gNg^-1$ s.t $ a = gn_1g^-1$, by group cancellation laws we have $n_1 = gag^-1$, since $a \in N$ we have that $n_1 \in gNg^-1$ since $n_1$ is arbitrary we have that $gNg^-1 = N$

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I don't know what RGR means but unless it means "abelian group" (in which case the statement is quite trivial) then

by group cancellation laws we have $n_1=gag^{-1}$

is incorrect. The correct equality is

$$n_1 = g^{-1}ag$$

so you can only deduce that $n_1\in g^{-1}Ng$ not $n_1\in gNg^{-1}$.

Finitness of the group is essential here. But you haven't used it in your proof. Let me give you an example that the original statement does not hold for infinite groups.

Let $\mathbb{F}_2 = <a, b>$ be a free group in 2 generators. Consider

$$h_n := a^{n}ba^{-n}$$ $$H_n := (h_n)$$

So $H_n$ is a subgroup generated by $h_n$. Now put

$$W_k := (H_k, H_{k+1}, H_{k+2}, \ldots)$$

So it's a subgroup generated by an infinite sequence of subgroups. Note that obviously

$$W_{k+1} \subseteq W_{k}$$

What is not trivial (but nevertheless true) is that this inclusion is proper, i.e. $W_{k+1} \neq W_{k}$. And since

$$aH_na^{-1}=H_{n+1}$$

then

$$aW_ka^{-1}=W_{k+1}$$

So concluding

$$aW_ka^{-1} \subset W_{k}$$ but $$aW_ka^{-1} \neq W_{k}$$

freakish
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  • Thanks for your answer, that was a really silly mistake to make on my part. – ಠ_ಠ Oct 26 '16 at 23:30
  • I was wondering how is $aH_na^{-1} = H_{n+1}$ since for e.g $a(a^{2n}b^2a^{-2})a^{-1} \ \not= (a^{n+1}ba^{-n-1}) ^2 $ – ಠ_ಠ Oct 26 '16 at 23:38
  • @JayK That's because for a generator you have $a(a^nba^{-n})a^{-1} = a^{n+1}ba^{-(n+1)}$, i.e. conjugation maps generator of $H_n$ to a generator of $H_{n+1}$. Elements of the form $a^{2n}b^2a^{-2}$ do not appear in any $H_n$, I'm not sure where did you get that from. Is that a typo? – freakish Oct 26 '16 at 23:51