Re: simple groups and permutation groups

In article <r875v3dqrd70pah649irfdsti105h613cl@xxxxxxx>,
quasi <quasi@xxxxxxxx> wrote:
On 1 Apr 2008 19:18:25 GMT, Marc Olschok <nobody@xxxxxxxxxxxxxxx>
wrote:

quasi <quasi@xxxxxxxx> wrote:
On 31 Mar 2008 19:31:09 GMT, Marc Olschok <nobody@xxxxxxxxxxxxxxx>
wrote:

quasi <quasi@xxxxxxxx> wrote:
On Mon, 31 Mar 2008 00:39:44 EDT, Jack Schmidt
<Jack.Schmidt.SciMath@xxxxxxxxx> wrote:

Let G be a simple group and let f : S_n --> G be a
surjective homomorphism for some positive integer n.

Why is G isomorphic to S_k, for some k =< n ?

Because G is cyclic of order two and n>=2.

Which would make it a trick question.

If that was really the wording of the assigned question then, while
technically not incorrect, I suspect the problem was posed in error.

Looks o.k. to me. The question is

Any simple epimorphic image of a permutation group S_n is
already isomorphic to some S_k.

Except that the only such simple epimorphic images are either the
trivial group or Z_2. Hence, unless it was an intentional trick
question, it seems liklely that the proposer of the problem was
confused about the set of possible isomorphism types.

Sorry, I had misread the text all the time by automatically
substituting "A_n" for "S_n". Does the question look more reasonable now?

I'd have to see the actual question.

Presumably:

If G is a finite simple group, and f:A_n --> G is a surjective
homomorphism, then G is A_k for some k<=n.

This is a bit better, since the homomorphic images of A_n for n=/=4
are A_n and {0}, so the only simple ones are A_n themselves when n=/=2
and n=/=4. And A_4 has a nontrivial homomorphic image, the quotient of
the Klein 4-group generated by the products of two disjoint
transpositions, which is isomorphic to Z/3Z, itself isomorphic to
A_3. So there is a bit more to do here.

--
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"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
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Arturo Magidin
magidin-at-member-ams-org

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