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Chief factors of a solvable group G, are they always normal in G? I know Chief factors are characteristically simple. Also, characteristic subgroups are normal in G. So I think Chief factors are normal in G. But I found an example, 1,V4,A4,S4 chief series, where Z2 is a chief factor, which is not normal in S4. Now I am confused.Please clarify this.

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    $\begingroup$ First: you probably mean "the factors that conform" a chief factor, as a chief factor itself is not even a subset of $\;G\;$, not to mention a subgroup or normal. Second, if you meant the above then, by definition, the answer is yes, as chief factors are formed from normal series, meaning each of the elements in the series is a normal subgroup of the big group. Do you have any other definition of "chief factor"? BTW, the group being solvable is irrelevant in this. $\endgroup$ – DonAntonio Mar 21 '17 at 11:45
  • $\begingroup$ @DonAntonio Thank you for the comment. Yes, chef factors are the quotients of the chief series. it is a subnormal series where series has no refinements without representation. But, what about the example for S4? $\endgroup$ – student Mar 21 '17 at 11:52
  • $\begingroup$ Well, in my definition, the chief factors are formed by the elements of a chief series, which is a normal, not subnormal, series. You can check this here: en.wikipedia.org/wiki/Chief_series , or here: books.google.co.il/… , or here: ms.uky.edu/~jack/2008-03-05-ChiefFactors.pdf $\endgroup$ – DonAntonio Mar 21 '17 at 12:10
  • $\begingroup$ Yes, I am sorry. it should be a normal series. But, still I have the confusion with the above example. weather chief factors are normal in G $\endgroup$ – student Mar 21 '17 at 12:15
  • $\begingroup$ Look, your question does not make sense. Only a subgroup of $G$ can be normal in $G$ or not normal in $G$, and the chief factors are not subgroups of $G$, they are quotient groups of subgroups, $\endgroup$ – Derek Holt Mar 21 '17 at 12:32
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I think your example is

$$1\lhd C_2\lhd V_4\lhd A_4\lhd G$$

But this is not a normal series since, as you mentioned, $\;C_2\rlap{\;\,/}\lhd S_4\;$ and thus we have no chief factors here.

Where we do have a normal series is with

$$1\lhd V_4\lhd A_4\lhd S_4\;\;\;(**)$$

and for this to be a chief series it must be that every factor has no subfactor that is normal in $\;S_4\;$ ...and this is true! The only non-trivial factor a quotient in the above series has is $\;C_2\lhd V_4\;$, but since $\;C_2\;$ is not normal in $\;S_4\;$ we don't care, and thus $\;(**)\;$ is a chief series ...

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  • $\begingroup$ Thank you very much. which means, Chief factors need not to be normal in the whole group. $\endgroup$ – student Mar 21 '17 at 12:32
  • $\begingroup$ It doesn't mean that. The statement "Chief factors need not to be normal in the whole group." does not make sense, because chief factors of a group are not subgroups of the group. $\endgroup$ – Derek Holt Mar 21 '17 at 12:35
  • $\begingroup$ @student Read the comment by Derek, which is the same I wrote in my first comment below your question: chief factors are different sets, let alone groups, than the whole group. In the above answer, $(**)$ is a chief series ( the elements of which are normal subgroups of $\;G=S_4\;$) , and the chief factors there are $\;V_4/1\cong V_4\,,\,\,A_4/V_4\cong C_3\,,\,\,S_4/A_4\cong C_2\;$ , which are not even subsets of $\;S_4\;$ ... $\endgroup$ – DonAntonio Mar 21 '17 at 12:46
  • $\begingroup$ Yes, I got it. Thank you very much. $\endgroup$ – student Mar 21 '17 at 13:12
  • $\begingroup$ @student Any time. $\endgroup$ – DonAntonio Mar 21 '17 at 13:14

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