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Problem 51 (Inverses In Groups)

Suppose that $G$ is a group with $a,b\in G$.

  1. Prove that the inverse of $a^{-1}$ is $a$.
  2. Prove that the inverse of $ab$ is $b^{-1}a^{-1}$.
  3. If $a_1,a_2,a_3,\ldots, a_n\in G$, state the inverse of $a_1a_2a_3\cdots a_n$. Use induction to prove your claim.
If you see yourself repeating an induction proof similar to what we've been doing, then you're on the right track.

Solution

Suppose $G$ is a group with $a,b\in G$. We must prove the inverse of $a^{-1}$ is $a$. We begin by letting $a\in G$. Since $G$ is a group we know $a$ has an inverse, which we will call $a^{-1}$. Since we have closure under the binary operation we know $a^{-1}\in G$. This means $a^{-1}$ has an inverse, which we will call $(a^{-1})^{-1}$. We must show $(a^{-1})^{-1} = a$.

Recall from the definition of inverse that $a^{-1} a = a a^{-1} = e$, where $e$ is the identity in $G$. We also know $(a^{-1})^{-1} a^{-1} = a^{-1} (a^{-1})^{-1} = e$. Since all of these are equal to $e$ we can say $$a^{-1} (a^{-1})^{-1} = a^{-1} a.$$ Since we have a binary operation we will multiply on the left by $a$ to get $$a(a^{-1} (a^{-1})^{-1}) = a(a^{-1} a).$$ By the associative property we know $$(a a^{-1} )(a^{-1})^{-1} = (a a^{-1}) a.$$ Applying the inverse we get $$ e(a^{-1})^{-1} = e a.$$ By the definition of the identity we now know $$ (a^{-1})^{-1} = a.$$

Next we will prove the inverse of $ab$ is $b^{-1}a^{-1}$. We compute $$\begin{align} (ab)(b^{-1}a^{-1} &= a(bb^{-1})a^{-1} &\text{associativity}\\ &= aea^{-1} &\text{inverse}\\ &=a^{-1} &\text{identity}\\ &=e &\text{inverse}. \end{align}$$

We claim the inverse of $a_1, a_2, a_3, \cdots, a_n \in G$ is $a_n^{-1}, a_{n-1}^{-1}, \cdots, a_2^{-1}, a_1^{-1}$. We will first prove the base case. Suppose that $n=1$. Then we have $a_n^{-1} = a_1^{-1}$. To check this we compute $$ \begin{align} aa^{-1} &= e &\text{inverse}. \end{align} $$ Thus we have proven the base case.

Now we suppose for some $k<n$ with $k\in \mathbb{N}$ that $(a_1 a_2 \cdots a_k)^{-1} = a_k^{-1}, a_{k-1}^{-1}, \cdots, a_1^{-1}.$ We compute $$\begin{align} (a_1 a_2 \cdots a_k a_{k+1})(a_{k+1}^{-1}, a_k^{-1}, \cdots, a_2^{-1}, a_1^{-1}) &= (a_1 a_2\cdots a_k)(a_{k+1} a_{k+1}^{-1})(a_k^{-1}, \cdots, a_2^{-1}, a_1^{-1}) &\text{associativity}\\ &=(a_1 a_2\cdots a_k)(e)(a_k^{-1}, \cdots, a_2^{-1}, a_1^{-1}) &\text{inverse}\\ &=(a_1 a_2\cdots a_k)(a_k^{-1}, \cdots, a_2^{-1}, a_1^{-1}) &\text{identity}\\ &= e &\text{inverse} \end{align}$$

Thus we have shown by mathematical induction that the inverse of $(a_1 a_2 \cdots a_n)$ is $(a_n^{-1} a_{n-1}^{-1} \cdots a_1^{-1})$. $\square$

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