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2

If $x\notin Z(G)$, then $C_G(x) \supsetneq Z(G)$ but $C_G(x) \subsetneq G$. Since $[G:Z(G)]=p^2$, we must have $[G:C_G(x)]=p$ for all conjugacy classes. So all conjugacy classes have $p$ elements, and since $p^4 = p^2 + kp,$ there must be $k = p^3-p$ nontrivial conjugacy classes, so $p^3+p^2-p$ overall (including $Z(G)$).


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In general, an action of a group $G$ on a set $X$ is a group homomorphism from $G$ to the group $S_X$ of permutations of the set $X$. This means that we send each group element $g \in G$ to some permutation of the elements in $X$, so each group element "acts" on $X$ by permuting its elements in some way. Usually, we write $g \cdot x$ to denote the element of ...


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What we are counting here is homomorphisms $\mathbb{Z}_5\to S_5$. The image of a generator of $\mathbb{Z}_5$ must have order dividing $5$, hence it is either the identity or a $5$-cycle. There are $\displaystyle\frac{5!}{5}=24$ $5$-cycles (the number of permutations of $5$ elements divided by the number of cyclic permutations) and one identity element, hence ...


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Hint: $A_n$ is the kernel of the signature morphism $$\epsilon:S_n\longrightarrow\lbrace-1,+1\rbrace$$ What can you say about the kernel of the composite $\epsilon\circ\psi$? ... Whence $\mathrm{Im}(\phi)\subset A_n$.


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I will show the one direction and the other is an exercises for you. Note that for a normal group $N$, $gN=Ng$ thus for $gn$ there is a uniqe $n_2$ such that $gn=n_2g$. Let $H$ be a normal subgroup of $G$ and $K=H\times H$ $|K:Stab(x)|$ is the size of orbit containing $x$. $$hxh'^{-1}=x$$ $$xh_2h'^{-1}=x$$ $$h_2=h'$$ Note that for any $h$, there is a ...


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We have the follwing nice theorem: Th.: Let $\;G\;$ be a finite $\;p$- group acting on a finite set $\;X\;$ , and put $$X^g:=\{x\in X\;;\;x^g=x\;\forall\;g\in G\}$$ Then, $$|X^G|=|X|\pmod p$$ Proof: We know $$X=\bigsqcup\,\mathcal Orb(x)$$ Now, $$|\mathcal Orb(x)|=[G:G_x]\;,\;\;G_x:=\{g\in G\;;\;x^g=x\}$$ and thus all the orbits are of size a ...


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Recall that a group action can be defined as follows. Let $X$ be a set and denote by $Sym(X)$ the group of all permutations of $X.$ Then an action of a group $G$ on the set $X$ is a group homomorphism $\varrho: G \rightarrow Sym(X).$ Then, the action $\varrho$ is called faithful if $ker\ \varrho = \{1_G\},$ as you noted in your question. Now, the group ...


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What conjugation by an element of $S_{n-1}$ preserves is the cycle structure plus the length of the cycle that position $n$ is part of. Thus each conjugacy class of $S_n$ splits into one class for each different length cycle in its cycle structure.


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Every group action of $G$ is uniquely a disjoint union of transitive group actions. These are all of the form $G/H$ where $H$ is a subgroup of $G$, and two conjugate subgroups give isomorphic group actions. So the problem reduces to classifying the conjugacy classes of subgroups of $G$, which is hard in general, but that's the general answer. The resulting ...


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I think you are right. The set $X$ is the set of all two element subsets of $G$. For the action of $G$ on $X$, first note that $G$ has a regular action on itself. Use this action to define an action on $X$. More precisely, if $g_i \mapsto h_i$ and $g_j \mapsto h_j$ under the regular action of $G$ on itself $(i \neq j)$, then the action of $G$ on $X$ is ...


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The two element subsets: $$\{0,1\},\{0,2\},\{0,3\},\{0,4\},\{0,5\},\{1,2\},\{1,3\},\{1,4\},\{1,5\},\{2,3\},\{2,4\},\{2,5\},\{3,4\},\{3,5\},\{4,5\}.$$ Left addition: Take $\left[a\right]_6 \in \mathbb{Z}_6$ and do $a+H$ (where $H$ is a two element subset of $G$). For example: $$1+\{0,1\} =\{1,2\}, 1+\{1,2\} = \{2,3\}, 1+\{2,3\} = \{3,4\}, 1+\{3,4\} = ...



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