# Prove that $g^{-1}hg$ is contained in set of injections with finite non-identities of a set onto itself

I'm struggling with a proof for a problem from Herstein's Abstract Algebra text. I'm very new to proof writing, so please bear with me. The problem is the following:

Let $S$ be an infinite set and let $M \subset A(S)$ (where $A(S)$ is the set of all 1-1 mappings of $S$ onto itself) be the set of all elements $f \in A(S)$ such that $f(s) \not= s$ for at most a finite number of $s \in S$. If $g \in A(S)$, prove that $g^{-1}Mg = \{ g^{-1}hg \mid h \in M \}$ must equal $M$.

In order to prove this, I need to show both that $g^{-1}Mg \subseteq M$ (that is, if $h \in M$, $g^{-1}hg \in M$) and that $M \subseteq g^{-1}Mg$. My attempt at showing that $g^{-1}Mg \subseteq M$ is the following:

Suppose that $h \in M$ moves only elements $s_1,s_2,...,s_m$, leaving the other elements of $S$ fixed. This means that for some element $s \not\in s_1,s_2,...s_m$, $h(s) = s$. Since $g \in A(S)$, we know that if $g(s_1) = s_2$, there exists some $g^{-1}$ such that $g^{-1}(s_2) = s_1$. Therefore, for all $g(s) \not\in s_1,s_2,...,s_m$, then $(g^{-1}hg)(s) = s$, meaning there is a finite number of $(g^{-1}hg)(s) \not = s$, so $g^{-1}hg \in M$.

The part I've really been struggling with is showing that $M \subseteq g^{-1}Mg$. I assume that I need to show that for some $h \in M$, $h \subseteq g^{-1}hg$, but I don't know how to go about doing this. I thought I could possibly show this by proving any $(s_1, s_2) \in h$ is also in $g^{-1}hg$, but it seems like my approach isn't working. It seems to me that $(g^{-1}hg)(s)$ could first transform $s$ to one of two places: either to an element such that $g(s) \in s_1,s_2,...s_m$, or to an element such that $g(s) \not\in s_1,s_2,...s_m$. If $g(s) \not\in s_1,s_2,...s_m$, then $h(g(s)) = g(s)$ and $g^{-1}(h(g(s))) = g^{-1}(g(s)) = s$. However, since we only know that $g(s) \not\in s_1,s_2,...,s_m$, we can't say that $s \not\in s_1,s_2,...,s_m$, so it's possible that $h(s) \not= s$, which doesn't seem to help me. How can I prove this set equivalence?

• Welcome to Math SX! There exists some (in the last but one paragraph)? – Bernard May 25 '17 at 19:39

Hint (for $M\subseteq g^{-1}Mg$):
If $f\in M$, $\;gfg^{-1}\in M$ by the reverse inclusion.