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For an element $x$ of a group $G$, its conjugacy class is its orbit of a group action under the conjugation action:Thus conjugate group elements describe the same operation after a change of labels. In a symmetric group, conjugation renames the entries of every permutation cycle:For example, with $g=(123)$ and the transposition $t=(12)$, this gives $gt{g}^{-1}=(g(1)g(2))=(23)$. Products use function composition of permutations, with the rightmost factor applied first. The complete six-element example is conjugacy classes of the symmetric group on three letters.

The conjugacy classes form a partition of a set of $G$, since conjugacy is an equivalence relation. In a symmetric group, two permutations lie in the same conjugacy class precisely when they have the same cycle type: renaming preserves the lengths of permutation cycles, and matching equal-length permutation cycles constructs a renaming in the other direction. This compresses many labelled permutations into a few structural types. See McKernan, Conjugation in Sn, lemma 6.2 and proposition 6.6.

The stabilizer subgroup of $x$ under the conjugation action is its centralizer $C_G(x)$. Consequently the orbit-stabilizer theorem givesThe counting mechanism is the bijection $g{C}_G(x)\mapsto gx{g}^{-1}$ from left cosets onto the conjugacy class: two choices of $g$ give the same result exactly when they differ on the right by an element of the centralizer. A large centralizer means many choices of labels leave the operation unchanged, hence a small conjugacy class. A conjugacy class has size one precisely when its element belongs to the center of a group. The class equation counts these orbits of a group action together.

If $y=gx{g}^{-1}$, cancellation of successive $g^{-1}g$ factors gives $y^k=g{x}^k{g}^{-1}$ for every positive integer $k$. Hence $y^k$ is the identity element exactly when $x^k$ is, proving that conjugate group elements have the same order of a group element. For permutations, the sign group homomorphism similarly givesThus conjugate permutations have the same sign of a permutation.

The converse is false: permutation order and sign do not determine conjugacy. In the symmetric group $S_6$, take $x=(123)$ and $y=(123)(456)$. Both have order three and positive sign. However, $x$ fixes three symbols whereas $y$ fixes none. If $x$ fixes a symbol $j$, then $gx{g}^{-1}$ fixes $g(j)$, and applying the inverse conjugation gives a bijection between the fixed-symbol sets. Their different fixed-point counts therefore prove that $x,y$ are not conjugate. Equivalently their cycle types $31^3$ and $3^2$ differ. Equal order and sign are necessary, but not sufficient, for conjugacy.

The conjugate group elements relation is $x\sim y$ when $y=gx{g}^{-1}$ for some $g\in G$. It is reflexive by taking the identity, symmetric because $x=g^{-1}yg$, and transitive because $y=gx{g}^{-1}$ and $z=hy{h}^{-1}$ imply $z=(hg)x(hg{)}^{-1}$. Hence it is an equivalence relation.

For a finite group, the centralizer $C_G(x)=\{g:gx=xg\}$ is a subgroup. The map $g\mapsto gx{g}^{-1}$ has equal values at $g,h$ exactly when $h^{-1}g\in C_G(x)$. Its fibers are therefore the left cosets of $C_G(x)$, givingBy Lagrange's theorem, this index divides $|G|$. This is also the orbit-stabilizer theorem for the conjugation action, with the centralizer as stabilizer.

False. Take the finite, hence countable, cyclic group $C_6=⟨a\mid a^6=1⟩$. Its elements $a^3$ and $a^2$ have orders two and three. A group embedding preserves the order of every element, whereas conjugate group elements have the same order: $(s^{-1}xs{)}^r=s^{-1}{x}^{r}s$. Therefore their images cannot be conjugate in any containing group. This gives the required obstruction. The torsion-free group hypothesis in Question 5 avoids this conflict of element orders.

A permutation cycle $(a_1{a}_2\cdots a_p)$ sends $a_1$ to $a_2$, each successive listed point to the next, and $a_p$ to $a_1$, fixing every unlisted point. Its length is $p$. For any permutation $h$,Thus conjugate group elements in the symmetric group have the same cycle length. Conversely, two cycles of the same length are conjugate: send their entries, in cyclic order, to one another and extend this bijection arbitrarily to the unused points. Length-one cycles represent the identity and cause no exception to this conclusion.

An odd-length permutation cycle is an even permutation, since its sign is $(-1{)}^{p-1}$. Let $c,d$ be such cycles on $p+2$ points and choose $h\in S_{p+2}$ with $hc{h}^{-1}=d$. If $h$ is odd, let $t$ be the transposition of the two points unused by the listed cycle $c$. Then $tc=ct$, and $ht$ is even withHence all the $p$-cycles are conjugate within $A_{p+2}$. The two spare letters supply the parity correction; they are the reason the same argument need not work with only one spare letter.

In $A_4$ the answer is no. For example, $(123)$ and $(132)$ are conjugated by the odd transposition $(23)$ in $S_4$. The centralizer of $(123)$ in $S_4$ is exactly its order-three cyclic subgroup: a commuting permutation fixes the unique fixed point $4$ and acts as a power of the cycle on the other three points. All these centralizing elements are even. Every other conjugating element differs from $(23)$ by a centralizing element, so it is also odd. There is no conjugator in $A_4$. Equivalently, the eight three-cycles split into two conjugacy classes of size $12/3=4$, illustrating the alternating conjugacy class splitting criterion.

Let $H(s,z)$ be the free homotopy from $\alpha$ to $\beta$ and put $v(s)=H(s,1)$. Because both endpoint loops are based at $x_0$, this is itself a loop based at $x_0$.

Cut the circle parameter at 1 and regard $H$ as a map on a square. Traversing its boundary gives the concatenation $\alpha \ast v\ast \bar{\beta }\ast \bar{v}$. The boundary is null-homotopic since the map extends over the square. Thus, in the fundamental group,Hence the corresponding elements are conjugate group elements. The moving basepoint path is precisely what distinguishes free homotopy from based homotopy.

Write $\bar{a}=f(a)$, $\bar{t}=f(t)$, and $\bar{b}={\bar{a}}^{\bar{t}}={\bar{t}}^{-1}\bar{a}\bar{t}$. The elements $\bar{a}$ and $\bar{b}$ are conjugate group elements, so they have the same order of a group element, say $n$. The defining relation becomesIterating conjugation gives ${\bar{b}}^j\bar{a}{\bar{b}}^{-j}={\bar{a}}^{2^j}$. Since ${\bar{b}}^n=1$, taking $j=n$ yields $\bar{a}={\bar{a}}^{2^n}$, and henceThe stated consequence of Fermat's little theorem implies $n=1$. Thus $\bar{a}=1$, and the surjective group homomorphism $f$ shows that $Q$ is generated by $\bar{t}$. Therefore $Q$ is a cyclic group.