# Show the set of inverse elements of a group is the same as the set of elements of the group

Let $L$ be a group. Define the set $$L^{-1}=\lbrace l^{-1}|l \in L \rbrace$$ Show that $L^{-1}=L$.

Let $l^{-1} \in L^{-1}$.Then $l \in L$. Since L is a group, $l^{-1} \in L$ , which implies that $L^{-1} \subset L$ . Another way I don know how to show.

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Note that $L$ is a group and for $l\in L$ the inverse $l^{-1}$ is by definition an element in $L$ such that $ll^{-1} = 1_L$. So by definition $L^{-1} \subseteq L$.

Let $l\in L$. Now $l^{-1}\in L$, so $l = (l^{-1})^{-1} \in L^{-1}$. So $L\subseteq L^{-1}$.

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If $l \in L$, then $l = (l^{-1})^{-1} \in L^{-1}$ because $l^{-1} \in L$.

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Recap of your work:

$(1)$ Since $L$ is a group, for each $l\in L$, its inverse $l^{-1}\in L^{-1}$ is necessarily in $L$, since $L$ is a group, and as such, for each $l \in L$, there exists a unique inverse $l^{-1} \in L$, meaning $ll^{-1} =l^{-1}l = e_L.\;$
So by definition,

$$L^{-1} \subseteq L.\tag{1}$$

That's half of the proof. Next, we show $L \subseteq L^{-1}$, and these two inclusions together then prove equality.

$(2)$ Let $l \in L$. Now, $l = (l^{-1})^{-1} \in L^{-1}$ since $l^{-1} \in L.$ Hence

$L \subseteq L^{-1}\tag{2}$

From $(1)$ together with $(2)$, it follows that $L^{-1} = L$.

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1) In a group every element has an inverse.

2) The inverse of a given element is unique.

3) Two different elements cannot have the same inverse since $g^{-1} = h^{-1}$ implies $g=h$.

These three things together tell you that the set of inverses of things in $L$ is the set $L$.

(Technically: If we consider the natural map $L \longrightarrow L^{-1}$ then 1 shows it is surjective, 2 shows it is well defined as a function, 3 shows that it is injective, hence it is a bijection of sets and since $L^{-1}\subseteq L$ it must be an equality).

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