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I'm reading the solution to one of my homework problems and am stuck on something. Here is the problem:

Let $N \leq G$ be a finite subgroup. Show that $gNg^{-1} \subseteq N$ if and only if $gNg^{-1} = N$.

The solution is to suppose that $gNg^{-1} \subseteq N$ and consider the map $\varphi: N \to gNg^{-1}$ defined by $\varphi(n) = gng^{-1}$. Observe that such a map is a bijection and thus $|N| = |gNg^{-1}|$. The equality then follows.

I'm just curious as to why we can assume that $\varphi$ is a bijection (i.e. why can't there exist $n_1, \, n_2 \in N$ such that $gn_1g^{-1} = gn_2g^{-1}$ where $n_1 \neq n_2$)?

Qiaochu Yuan
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Ryan
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    It isn't assuming any such thing. It is telling you that it is true and you should think about it and prove it. Omnom's answer tells you how to resolve your other issue. – zibadawa timmy Oct 28 '14 at 07:06

2 Answers2

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The fact that $\varphi$ is a bijection follows from the invertibility of group multiplication.

Start with $gn_1g^{-1} = gn_2g^{-1}$. First multiply both sides on the left by $g^{-1}$. Then multiply on the right by $g$.

user1729
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Ben Grossmann
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Another way of proving the result is as follows:

$gNg^{-1}\subseteq N\Rightarrow gN\subseteq Ng$. However, cosets all have equal cardinality (whether they are left or right cosets), so as we are in a finite group we have that $gN=Ng$, and so $gNg^{-1}=N$ as required.

Notes:

  1. It is tempting to go $gN\subseteq Ng$, and cosets are disjoint of equal...but left cosets are disjoint and equal, and right cosets are disjoint or equal, but we cannot say the same about comparing a left coset with a right coset.
  2. The fact that we are in a finite group is important - the result does not hold for infinite groups. See the answers to this question.
Community
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user1729
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