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Is $Z( G_1 \oplus G_2 \oplus G_3 \oplus\cdots\oplus G_n) = Z(G_1) \oplus Z (G_2) \oplus Z(G_3) \oplus \cdots \oplus Z$ $(G_n)$ true, where $ G_1, G_2, G_3 \cdots G_n$ are finite groups?$Z$,here refers to center of group.

What is the condition for the validity of this statement?

This question arose when I was dealing with question of finding the conjugacy classes & their sizes of $ \mathbb Z_2\oplus S_3$.

Also, please check my approach for finding the conjugacy classes & their sizes of $ D_8$.

$Z(D_8)=\{1,r^2\}$

$C_{D_8}(r)=\{1,r^2,r,r^3\}$

$C_{D_8}$($r^3$)={1,$r^2$,$r$,$r^3$}

$C_{D_8}$(s)={1,$r^2$,$r$,$sr^2$}

$C_{D_8}$($sr$)={1,$r^2$,$sr$,$sr^3$}

$C_{D_8}$($sr^2$)={1,$r^2$,$sr^2$,$s$}

$C_{D_8}$($sr^3$)={1,$r^2$,$sr^3$,$sr$}

  • Does here order of elements in the centralizer matter?

From the above list we get,

$C_{D_8}$($r$)=$C_{D_8}$($r^3$);$C_{D_8}$($s$)=$C_{D_8}$($sr^3$);$C_{D_8}$($sr$)=$C_{D_8}$($sr^3$)

Here I've doubt-

The conjugacy class of $r$ and $r^3$={$r$,$r^3$}. The conjugacy class of $s$ and $sr^2$={$s$,$sr^2$}. The conjugacy class of $sr$ and $sr^3$={$sr$,$sr^3$}. Each conjugacy class is obtained by removing the centere of $D_8$ from the centralizer of an element of the conjugacy class.

  • Is it a general method of finding the conjugacy class of any element by first finding the centralizer of given element and then removing the centre?

When I applied the same procedure with same question for $\mathbb Z_2$$\oplus$$S_3$

$\mathbb Z_2$=${0,1}$ & $S_3$=${(1),(132),(213),(231),(312),(321)}$

then $ \mathbb Z_2\ $$\oplus $ $S_3$=${(0,(1)),(0,(132)),(0,(213)),(0,(231)),(0,(312)),(0,(321)),(1,(1)),(1,(132)),(1,(213)),(1,(231)),(1,(312)),(1,(321))}$

$C_{ \mathbb Z_ 2\oplus S_3}$($0,(132)$)=${(0,(1)),(0,(132)),(0,(213)),(0,(231)),(0,(312)),(0,(321)),(1,(1)),(1,(132)),(1,(213)),(1,(231)),(1,(312)),(1,(321))}$=$\mathbb Z_2$$\oplus$$S_3$

Which implies that $\mathbb Z_2$$\oplus$$S_3$ is an abelian group,but it is not,as $S_3$ is a non-abelian group(WHERE I COMMITED MISTAKE?).

Also,give me some hint to prove-" The conjugacy classes for the direct product of two groups comes directly (via cartesian product) of the conjugacy classes of the two groups"(this i got from https://uk.answers.yahoo.com/question/index?qid=20140308110451AAaWREE)

Thanks for giving your time to my problem!!

Any suggestions are heartly welcome!

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  • $\begingroup$ I gave you, in a comment on your comment on my answer to another question, a simple example where the conjugacy class is not obtained by removing the center from the centralizer. $\endgroup$ – Gerry Myerson Jun 24 '16 at 7:27
  • $\begingroup$ Note: "order" means something special in group theory. When you ask about the "order of elements in the centralizer", it's not clear to me whether you are asking about their order in the group theory sense, or just asking about the order in which you list them. $\endgroup$ – Gerry Myerson Jun 24 '16 at 7:29
  • $\begingroup$ @GerryMyerson:I apologize for confusion,from order-i mean order in which we list elements. $\endgroup$ – P.Styles Jun 24 '16 at 7:34
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    $\begingroup$ @PKStyles : When I mentioned the difference in appearance between {1,$r^2$, $r$, $r^3$} and $\{1,r^2,r,r^3\}$ I was talking about typography, not about the mathematical content. The first has a mismatch in font sizes between the {braces} and the other characters, and between the digit $1$ and the other characters. The second doesn't. Putting the whole thing between a single pair of dollar signs assures not only that that problem is avoided but also that proper spacing between characters is used. $\qquad$ $\endgroup$ – Michael Hardy Jun 24 '16 at 16:06
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    $\begingroup$ Notice, for example, that proper spacing is used in $C_{D_8}(r)=\{1,r^2,r,r^3\}$ and not in $C_{D_8}$($r^3$)={1,$r^2$,$r$,$r^3$}. In the latter the $\text{“}{=}\text{''}$ is smaller than in the former and is too close to the things preceding and following it. Besides, putting the whole thing in jut one pair of dollar signs is SIMPLER. So you're making the code pointlessly complicated and getting inferior typography. $\qquad$ $\endgroup$ – Michael Hardy Jun 24 '16 at 16:10
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Is $Z(G_1\oplus\ldots\oplus G_n)=Z(G_1)\oplus\ldots\oplus Z(G_n)$?

Yes, since $$\begin{eqnarray*} (g_1,\ldots,g_n)\in Z(G_1\oplus\ldots\oplus G_n)\hspace{-60mm}\\ &\Leftrightarrow&(g_1,\ldots,g_n)(h_1,\ldots,h_n)=(h_1,\ldots,h_n)(g_1,\ldots,g_n)\text{ for all }(h_1,\ldots,h_n)\\ &\Leftrightarrow&(g_1h_1,\ldots,g_nh_n)=(h_1g_1,\ldots,h_ng_n)\text{ for all }(h_1,\ldots,h_n)\\ &\Leftrightarrow&g_ih_i=h_ig_i\text{ for all }i,\;h_i\in G_i\\ &\Leftrightarrow&g_i\in Z(G_i)\text{ for all }i\\ &\Leftrightarrow&(g_1,\ldots,g_n)\in Z(G_1)\oplus\ldots\oplus Z(G_n) \end{eqnarray*}$$

Does here order of elements in the centralizer matter?

No, they are just sets. However note that some of them don't seem correct, eg $C_{D_8}(r^3)=\{1,r,r^2,r^3\}$, not $\{1,r^2,s,r^3\}$.

Is it a general method of finding the conjugacy class of any element by first finding the centralizer of given element and then removing the centre?

No. Eg consider $D_{10}=\langle r,s\mid r^5=s^2=(rs)^2=1\rangle$. Then $Z(D_{10})=\{1\}$, $C_{D_{10}}(s)=\{1,s\}$ but the conjugacy class of $s$ is $\{s,rs,r^2s,r^3s,r^4s\}$.

(Implicit question) What is a general method for finding conjugacy classes in a finite group?

To find the conjugacy class of $g\in G$, compute $hgh^{-1}$ for each $h\in G$ (removing any duplicates). It's possible to be more efficient, but this will give the correct answer.

Your last question needs some more details. What is $sr^3$? What was your reasoning to determine $C_{\mathbb Z_2\oplus S_3}(sr^3)$?

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  • $\begingroup$ :Thanks,for clearing my doubts!!.Actually,in place of $sr^3$ its $(0,(132))$.(i've just edited it.) $\endgroup$ – P.Styles Jun 24 '16 at 6:42
  • $\begingroup$ :but,while finding normalizer,the order of elements matter(in which they are listed)? – $\endgroup$ – P.Styles Jun 24 '16 at 7:49
  • $\begingroup$ can someone give me some hint to prove the last highlighted statement. $\endgroup$ – P.Styles Jun 24 '16 at 20:20

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