Recipe

Compute a conjugate ghg⁻¹

Use when: you're given specific elements $g$ and $h$ of a group and asked to compute the single element $ghg^{-1}$

Goal: the element $ghg^{-1}$ , simplified to the group's own standard notation

Taught in Conjugation

Quick recipe

Find $g^{-1}$
Compute $ghg^{-1}$
Check the order is preserved

1
Write down the inverse of the conjugating element first. $g^{-1}$ is needed on the right side of $ghg^{-1}$ no matter which method Step 2 uses.

Example

In $S_4$ , let $g=(1\,4)$ . A transposition is its own inverse, so $g^{-1}=(1\,4)$ too.

Why: $g^{-1}$ appears explicitly in the definition of a conjugate, so the product can't even be written down without it.
2
Multiply the three elements together in the fixed order $g$ , then $h$ , then $g^{-1}$ , or use a shortcut when one applies.

Example

Continuing with $g=(1\,4)$ and $h=(1\,2\,3)$ in $S_4$ : trace each point, applying $g^{-1}$ first, then $h$ , then $g$ (rightmost acts first). For $x=2$ : $g^{-1}$ fixes $2$ , $h$ sends $2\to3$ , $g$ fixes $3$ , so $2\to3$ . Repeating for $x=1,3,4$ gives $1\to1$ , $3\to4$ , $4\to2$ . $ghg^{-1} = (2\ 3\ 4)$
- $h$ is written in cycle notation and the group is some $S_n$ — rename every entry of $h$ 's cycles using $g$ (the shortcut below).
- the group isn't a symmetric group, or the elements are given some other way — compose the three elements directly, one multiplication at a time, using whatever relations the group provides.
⚡ Shortcut

Rename every entry of $h$ 's cycle notation using $g$ instead: $g(1)=4$ , $g(2)=2$ , $g(3)=3$ , so $(1\ 2\ 3)$ becomes $(4\ 2\ 3)$ . That's the same cycle as $(2\ 3\ 4)$ written from a different starting point. It matches the direct computation with no tracing required.

Why: this is exactly the definition of conjugate applied to three specific elements. It can be computed only two ways: composing directly, or, inside $S_n$ , by swapping names.
3
Confirm the computed conjugate has the same order as $h$ . A mismatch means Step 2 has an arithmetic error, not that the theorem failed.

Example

$h=(1\,2\,3)$ has order $3$ . The computed $ghg^{-1}=(2\,3\,4)$ is also a 3-cycle, order $3$ , matching as guaranteed.

Why: conjugation always preserves order, as proved above, so this check catches a slip before it carries into later work.

Common mistakes

Computing $ghg$ instead of $ghg^{-1}$ , forgetting the final inverse.
Composing the three elements in the wrong order, since $ghg^{-1}$ reads right-to-left like ordinary function composition.
Using the cycle-rename shortcut outside a symmetric group, where it doesn't apply at all.
Trusting the shortcut's answer without checking the resulting cycle has the same length as the original.