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Oct 5, 2015 at 11:04 comment added Behsa I am sorry I mean about a new character degree in one of them which does not exist in other one.
Oct 5, 2015 at 10:59 comment added Derek Holt What do you mean by "this group"? All of the groups we have been discussing have order $3|{\rm PGL}(2,q)|$ so the character degrees will certainly be different from those of ${\rm PGL}(2,q)$.
Oct 5, 2015 at 10:55 comment added Behsa Is there any difference between the character degrees of ${\rm PGL}(2,q)$ and this group?
Oct 5, 2015 at 10:53 vote accept Behsa
Oct 5, 2015 at 10:48 history edited Derek Holt CC BY-SA 3.0
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Oct 5, 2015 at 9:04 comment added Behsa I am very thankful for the complete and very useful comments.
Oct 5, 2015 at 8:59 history edited Derek Holt CC BY-SA 3.0
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Oct 5, 2015 at 8:56 comment added Behsa I am sorry for this mistake. I consider that the Schur Multiplier is equal to 2. And for the second group in the above discussion we have $G\cong S_3\times {\rm PSL}(2,q)$, is this true?
Oct 5, 2015 at 8:40 comment added Geoff Robinson Note that $q = 9$ is a case where there is already non-split extension of $\mathbb{Z}/3\mathbb{Z}$ by ${\rm PSL}(2,9)$ ( I mean by this a perfect triple cover of ${\rm PSL}(2,9)$).
Oct 5, 2015 at 8:26 history answered Derek Holt CC BY-SA 3.0