What is the inverse cycle of permutation?

Given cyclic permutations, for example,

$$σ = (123)$$, $$σ_{2} = (45)$$,

what are the inverse cycles $$σ^{-1}$$, $$σ_2^{-1}$$?

Regards.

• To find the inverse of a permutation just write it backwards. If $\tau = (1243)(67)$ then $\tau^{-1}=(76)(3421)$ [which can then be rewritten as $\tau^{-1}=(1342)(67)$ ] – Bill Cook Mar 21 '12 at 15:46
• @BillCook : Thanks , can you give me a hint regarding how to prove it ? – JAN Mar 21 '12 at 15:49

To elaborate on my comment:

To find the inverse of a permutation just write it backwards. If $\tau = (1243)(67)$ then $\tau^{-1}=(76)(3421)$ which can then be rewritten as $\tau^{-1}=(1342)(67)$.

For for the above question: $(123)^{-1}=(321)=(132)$.

How does one prove this?

First consider a single cycle: $\sigma=(a_1a_2\dots a_k)$. If $b \not\in \{a_1,\dots,a_k\}$, then $\sigma(b)=b$ so $\sigma^{-1}(b)=b$. Thus $b$ shouldn't appear in the inverse. Next $\sigma(a_i)=a_{i+1}$ so $\sigma^{-1}(a_{i+1})=a_i$.
Thus if $\sigma$: $a_1 \mapsto a_2 \mapsto a_3 \mapsto \cdots \mapsto a_k \mapsto a_1$,
then $\sigma^{-1}$: $a_k \mapsto a_{k-1} \mapsto a_{k-2} \mapsto \cdots \mapsto a_1 \mapsto a_k$.
This is precisely the cycle $(a_k,a_{k-1}\dots,a_2, a_1)$ which is nothing more than $\sigma$ written backwards.

Now what about a list of cycles? Say $\sigma=\sigma_1\cdots \sigma_\ell$. Recall that $\sigma^{-1}=(\sigma_1\cdots \sigma_\ell)^{-1}=\sigma_\ell^{-1}\cdots \sigma_1^{-1}$. So we reverse the list of cycles and then write each one backwards -- thus the inverse is just the whole thing written backwards.

One thing to note: This still works even if $\sigma$ is not written in terms of disjoint cycles.

• $\sigma^{-1}=(\sigma_1\cdots \sigma_\ell)^{-1}=\sigma_\ell^{-1}\cdots \sigma_1^{-1}$. Do we need to reverse the order of these cycles? I think the multiplication of the cycles are commutative. – user398843 Jan 26 '19 at 20:40
• If they are disjoint cycles, they commute and the order doesn't have to be reversed. However, if the cycles aren't disjoint (like the permutation hasn't been "simplified"), they need to be reversed. – Bill Cook Jan 26 '19 at 22:01

So, you have that $\sigma_1$ is the cycle such that,

\begin{align} 1 \mapsto 2 \\ 2 \mapsto 3 \\ 3 \mapsto 1\end{align}

It's inverse, $\sigma_1^{-1}$ is a cycle such that composition, $\sigma_1 \circ \sigma_1^{-1}=\sigma_1^{-1} \circ \sigma_1$ is identity. So, the inverse cycle should look like :

\begin{align} 2 \mapsto 1 \\ 3 \mapsto 2 \\ 1 \mapsto 3\end{align}

What is this in cycle notation?

$\sigma_1^{-1} \equiv(213)$

I'll let you try the other one.

A particularly easy way of doing this, once you understand what the inverses do is: just to write the cycle backwards!

Note that for $(123)$, this is just $(321)$. Now, recall, that set of all permutations form a group. In a group, inverses are unique. So, can you tell me why $(321)$ and $(213)$ are the same?

• Got it , then the inverse of $(123)$ is $(321)$ – JAN Mar 21 '12 at 15:53
• Right. See my edit as well. – user21436 Mar 21 '12 at 16:00

Another way to look at Bill Cook's fifth paragraph in his answer:

First consider a single cycle: $\sigma=(a_1a_2\dots a_k)$. If $b \not\in \{a_1,\dots,a_k\}$, then $\sigma(b)=b$ so $\sigma^{-1}(b)=b$. Thus $b$ shouldn't appear in the inverse. Next $\sigma(a_i)=a_{i+1}$ so $\sigma^{-1}(a_{i+1})=a_i$. Thus if $\sigma$: $a_1 \mapsto a_2 \mapsto a_3 \mapsto \cdots \mapsto a_{k - 1} \mapsto a_k \mapsto a_1$, then $\sigma^{-1}$:
$a_1 \color{aqua}{\leftarrow} a_2 \color{aqua}{\leftarrow} a_3 \color{aqua}{\leftarrow}\cdots \color{aqua}{\leftarrow} a_{k - 1} \color{aqua}{\leftarrow} a_k \color{aqua}{\leftarrow} a_1 \iff$
$a_k \color{aqua}{\mapsto} a_{k-1} \color{aqua}{\mapsto} a_{k-2} \color{aqua}{\mapsto} \cdots \color{aqua}{\leftarrow} a_2 \color{aqua}{\mapsto} a_1 \color{aqua}{\mapsto} a_k$.
This is precisely the cycle $(a_k,a_{k-1}\dots,a_2, a_1)$ which is nothing more than $\sigma$ written backwards.

Every permutation n>1 can be expressed as a product of 2-cycles.

And every 2-cycle (transposition) is inverse of itself.

Therefore the inverse of a permutations is Just reverse products of its 2-cycles

(ab)^-1 = b^-1 a^-1