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Can someone please verify my answers to the following questions?

Answer true or false to the following questions:

  1. Two elements of a group in the same conjugacy class must have the same order

  2. A group of order 24 can have 5 conjugacy distinct classes of cardinalities 1, 4,4,6, and 12 respectively.

  3. The group $S_3$ has three conjugacy classes, of cardinalities 1, 2, and 3, respectively.

  4. An element is in the center of a group $G$ if and only if its centralizer is all of $G$

  5. Every group has at least one conjugacy class consisting of only one element

  6. If $H$ is a normal subgroup of $G$, then it is stable under the action of conjugation on $G$.

  7. The group $\mathbb{Z}_{17} \times \mathbb{Z}_2$ has 34 distinct conjugacy classes

  8. In any finite group $G$, the order of the centralizer of any element divides $|G|$.

  9. In an abelian group, the centralizer of each element is trivial.

  10. An abelian subgroup of a group is always normal

  1. True. Let $x \in G$. Consider the element $gxg^{-1}$, where $g \in G$. Let $n$ be the order of $x$. Then, we have $(gxg^{-1})^n = g^nxg^{-n} = g^n g^{-n} = e$. So, we have just shown that the order of $y$ is less than or equal to the order of $x$ if $y$ is conjugate to $x$. But this implies that the orders of $x$ and $y$ are equal, since $x$ is conjugate to $y$ if and only if $y$ is conjugate to $x$ (and so the inequality holds both ways, so the orders must be equal).

  2. False, since $1+4+4+6+12 \neq 24$

  3. True. The conjugacy classes are $\{ e \}$, $\{ (1 2 3), (1 3 2) \}$, $\{ (1 2), ( 1 3), (2 3) \}$.

  4. True. This is easy to see from the definitions.

  5. True. The identity element comprises a conjugacy class consisting of only one element.

  6. True, since $gHg^{-1} = H$, by definition.

  7. True. The group $\mathbb{Z}_{17} \times \mathbb{Z}_2$ is abelian. Therefore, each element comprises a distinct conjugacy class. Since the order of the group is 34, there are 34 distinct conjugacy classes.

  8. True. The centralizer of any element in a group $G$ forms a subgroup of $G$. Therefore, by Lagrange's theorem, the order of the centralizer divides $|G|$.

  9. False. In an abelian group, the centralizer of each element is the entire group.

  10. False. I can't seem to find a counter-example, but my intuition tells me that the statement is incorrect. Can someone please let me know of a counterexample?

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closed as too broad by user147263, muaddib, user91500, Claude Leibovici, Mike Pierce Jul 19 '15 at 4:36

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ By writing $(gxg^{-1})^n=g^nx^ng^{-n}$ you're assuming your group is abelian. $\endgroup$ – coldnumber Jul 18 '15 at 21:13
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    $\begingroup$ For a counter-example to 10, consider the group generated by the cycle $(12)$ in $S_3$. Similarly that generated by $(1234)$ in $S_4$. $\endgroup$ – john Jul 18 '15 at 21:16
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    $\begingroup$ Number one should be $(gxg^{-1})^{n}=gx^{n}g^{-1}$. Since, for instance, $(gxg^{-1})^2 = gxg^{-1}gxg^{-1}=gx^{2}g^{-1}$. By induction this generalizes. $\endgroup$ – TuoTuo Jul 18 '15 at 21:18
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    $\begingroup$ It is not really a good practice to upload such question with lots of subquestions. If you have a doubt about the correctness of either solution you could ask about it, but some of the questions are quite straightforward and it seems you don't really have a problem with them. $\endgroup$ – Pedro Tamaroff Jul 18 '15 at 21:34
  • $\begingroup$ @PedroTamaroff I apologize for that. $\endgroup$ – Lord Varys Jul 18 '15 at 21:38
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For $10$, take any group $A_n$ with $n\ge 5$. It is well-known that such groups are simple, that is, they have no non-trivial, normal subgroups. But the subgroup generated by $(1,2)$ is certainly abelian.

I find the rest of answers ok.

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