The probability that two elements of a finite nonabelian simple group commute

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It is mentioned in here (last paragraph of the first page) that Dixon proved the following result: the probability that two elements of a finite nonabelian simple group commute is at most $frac112$. I cannot seem to find this proof anywhere online. Do you know a proof of this fact? Or do you have a reference for it? (I am actually more interested in the original proof)

edited Sep 19 at 5:20

Yemon Choi

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user129021

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In fact, for any $epsilon$, there will be at most finitely many simple groups $G$ with $cp(G) geq epsilon$. This follows from Dixon's argument (linked below) and the fact that for any $n$ there are at most finitely many finite simple groups with an irreducible representation of dimension $n$. See e.g. mathoverflow.net/a/27365/1345
â€“Â Ian Agol
Sep 19 at 4:07

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edited Sep 19 at 5:20

Yemon Choi

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asked Sep 19 at 3:21

user129021

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In fact, for any $epsilon$, there will be at most finitely many simple groups $G$ with $cp(G) geq epsilon$. This follows from Dixon's argument (linked below) and the fact that for any $n$ there are at most finitely many finite simple groups with an irreducible representation of dimension $n$. See e.g. mathoverflow.net/a/27365/1345
â€“Â Ian Agol
Sep 19 at 4:07

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edited Sep 19 at 5:20

Yemon Choi

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asked Sep 19 at 3:21

user129021

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reference-request gr.group-theory pr.probability finite-groups

edited Sep 19 at 5:20

Yemon Choi

16.2k54699

asked Sep 19 at 3:21

user129021

361

edited Sep 19 at 5:20

Yemon Choi

16.2k54699

asked Sep 19 at 3:21

user129021

361

edited Sep 19 at 5:20

Yemon Choi

16.2k54699

edited Sep 19 at 5:20

Yemon Choi

16.2k54699

edited Sep 19 at 5:20

Yemon Choi

16.2k54699

asked Sep 19 at 3:21

user129021

361

asked Sep 19 at 3:21

user129021

361

asked Sep 19 at 3:21

user129021

361

2

In fact, for any $epsilon$, there will be at most finitely many simple groups $G$ with $cp(G) geq epsilon$. This follows from Dixon's argument (linked below) and the fact that for any $n$ there are at most finitely many finite simple groups with an irreducible representation of dimension $n$. See e.g. mathoverflow.net/a/27365/1345
â€“Â Ian Agol
Sep 19 at 4:07

add a commentÂ |Â

2

In fact, for any $epsilon$, there will be at most finitely many simple groups $G$ with $cp(G) geq epsilon$. This follows from Dixon's argument (linked below) and the fact that for any $n$ there are at most finitely many finite simple groups with an irreducible representation of dimension $n$. See e.g. mathoverflow.net/a/27365/1345
â€“Â Ian Agol
Sep 19 at 4:07

In fact, for any $epsilon$, there will be at most finitely many simple groups $G$ with $cp(G) geq epsilon$. This follows from Dixon's argument (linked below) and the fact that for any $n$ there are at most finitely many finite simple groups with an irreducible representation of dimension $n$. See e.g. mathoverflow.net/a/27365/1345
â€“Â Ian Agol
Sep 19 at 4:07

add a commentÂ |Â

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You can find Dixon's argument here (Google Books). (J.D..Dixon, Solution to Problem 176, Canadian Mathematical Bulletin 16 (1973), p302.)

edited Sep 19 at 6:58

YCor

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answered Sep 19 at 3:45

Keith Kearnes

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Note that Dixon's argument can be simplified a little. No finite simple group can have an irreducible complex character of degree $2$. If $G$ were a simple group with such a representation, then $|G|$ would be even (since irreducible character degrees divide the group order). Then $G$ would contain an involution $t$ by Cauchy's theorem, and $t$ would be in $Z(G)$ since $t$ would be represented by $-I$ (on consideration of determinant), contrary to simplicity.
â€“Â Geoff Robinson
Sep 19 at 10:48

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up vote
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You can find Dixon's argument here (Google Books). (J.D..Dixon, Solution to Problem 176, Canadian Mathematical Bulletin 16 (1973), p302.)

edited Sep 19 at 6:58

YCor

25.4k277120

answered Sep 19 at 3:45

Keith Kearnes

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Note that Dixon's argument can be simplified a little. No finite simple group can have an irreducible complex character of degree $2$. If $G$ were a simple group with such a representation, then $|G|$ would be even (since irreducible character degrees divide the group order). Then $G$ would contain an involution $t$ by Cauchy's theorem, and $t$ would be in $Z(G)$ since $t$ would be represented by $-I$ (on consideration of determinant), contrary to simplicity.
â€“Â Geoff Robinson
Sep 19 at 10:48

add a commentÂ |Â

up vote
7
down vote

You can find Dixon's argument here (Google Books). (J.D..Dixon, Solution to Problem 176, Canadian Mathematical Bulletin 16 (1973), p302.)

edited Sep 19 at 6:58

YCor

25.4k277120

answered Sep 19 at 3:45

Keith Kearnes

5,25912438

3

Note that Dixon's argument can be simplified a little. No finite simple group can have an irreducible complex character of degree $2$. If $G$ were a simple group with such a representation, then $|G|$ would be even (since irreducible character degrees divide the group order). Then $G$ would contain an involution $t$ by Cauchy's theorem, and $t$ would be in $Z(G)$ since $t$ would be represented by $-I$ (on consideration of determinant), contrary to simplicity.
â€“Â Geoff Robinson
Sep 19 at 10:48

add a commentÂ |Â

up vote
7
down vote

You can find Dixon's argument here (Google Books). (J.D..Dixon, Solution to Problem 176, Canadian Mathematical Bulletin 16 (1973), p302.)

edited Sep 19 at 6:58

YCor

25.4k277120

answered Sep 19 at 3:45

Keith Kearnes

5,25912438

You can find Dixon's argument here (Google Books). (J.D..Dixon, Solution to Problem 176, Canadian Mathematical Bulletin 16 (1973), p302.)

edited Sep 19 at 6:58

YCor

25.4k277120

answered Sep 19 at 3:45

Keith Kearnes

5,25912438

edited Sep 19 at 6:58

YCor

25.4k277120

edited Sep 19 at 6:58

YCor

25.4k277120

edited Sep 19 at 6:58

YCor

25.4k277120

answered Sep 19 at 3:45

Keith Kearnes

5,25912438

answered Sep 19 at 3:45

Keith Kearnes

5,25912438

answered Sep 19 at 3:45

Keith Kearnes

5,25912438

3

Note that Dixon's argument can be simplified a little. No finite simple group can have an irreducible complex character of degree $2$. If $G$ were a simple group with such a representation, then $|G|$ would be even (since irreducible character degrees divide the group order). Then $G$ would contain an involution $t$ by Cauchy's theorem, and $t$ would be in $Z(G)$ since $t$ would be represented by $-I$ (on consideration of determinant), contrary to simplicity.
â€“Â Geoff Robinson
Sep 19 at 10:48

add a commentÂ |Â

3

Note that Dixon's argument can be simplified a little. No finite simple group can have an irreducible complex character of degree $2$. If $G$ were a simple group with such a representation, then $|G|$ would be even (since irreducible character degrees divide the group order). Then $G$ would contain an involution $t$ by Cauchy's theorem, and $t$ would be in $Z(G)$ since $t$ would be represented by $-I$ (on consideration of determinant), contrary to simplicity.
â€“Â Geoff Robinson
Sep 19 at 10:48

Note that Dixon's argument can be simplified a little. No finite simple group can have an irreducible complex character of degree $2$. If $G$ were a simple group with such a representation, then $|G|$ would be even (since irreducible character degrees divide the group order). Then $G$ would contain an involution $t$ by Cauchy's theorem, and $t$ would be in $Z(G)$ since $t$ would be represented by $-I$ (on consideration of determinant), contrary to simplicity.
â€“Â Geoff Robinson
Sep 19 at 10:48

add a commentÂ |Â

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