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Pete L. Clark
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Why are the Sporadic Simple Groupssporadic simple groups HUGE?

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Alon Amit
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I'm merely a grad student right now, but I don't think an exploration of the sporadic groups is standard fare for graduate algebra, so I'd like to ask the experts on MO. I did a little reading on them and would like some intuition about some things.

For example, the order of the monster group is over $8\times 10^53$$8\times 10^{53}$, yet it is simple, so it has no normal subgroups...how? What is so special about the prime factorization of its order? Why is it $2^{46}$ and not $2^{47}$? Why is it not possible to extend it to obtain that additional power of 2 without creating a normal subgroup? Some of the properties seem really arbitrary, and yet must be very fundamental to the algebra of groups.

I don't think I'm the only person curious about this, but I hesitated posting due to my relative inexperience.

I'm merely a grad student right now, but I don't think an exploration of the sporadic groups is standard fare for graduate algebra, so I'd like to ask the experts on MO. I did a little reading on them and would like some intuition about some things.

For example, the order of the monster group is over $8\times 10^53$, yet it is simple, so it has no normal subgroups...how? What is so special about the prime factorization of its order? Why is it $2^{46}$ and not $2^{47}$? Why is it not possible to extend it to obtain that additional power of 2 without creating a normal subgroup? Some of the properties seem really arbitrary, and yet must be very fundamental to the algebra of groups.

I don't think I'm the only person curious about this, but I hesitated posting due to my relative inexperience.

I'm merely a grad student right now, but I don't think an exploration of the sporadic groups is standard fare for graduate algebra, so I'd like to ask the experts on MO. I did a little reading on them and would like some intuition about some things.

For example, the order of the monster group is over $8\times 10^{53}$, yet it is simple, so it has no normal subgroups...how? What is so special about the prime factorization of its order? Why is it $2^{46}$ and not $2^{47}$? Why is it not possible to extend it to obtain that additional power of 2 without creating a normal subgroup? Some of the properties seem really arbitrary, and yet must be very fundamental to the algebra of groups.

I don't think I'm the only person curious about this, but I hesitated posting due to my relative inexperience.

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Pete L. Clark
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REDace0
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