The Sylow Theorems and Groups of Small Order
Subgroups of prime-power order exist, are all conjugate, and their number is pinned down by two arithmetic conditions. That is enough to rule out simple groups of many orders and to finish the classification of small ones.
What a learner can do afterwards
- Count the possible Sylow subgroups from the congruence and divisibility conditions
- Show that no group of a given order can be simple
- List the groups of a small order and justify that the list is complete
1 · Read
You start with a finite group and a prime p. A Sylow p-subgroup is a subgroup whose order is the largest power of p that divides the group order. Write the order as p raised to a times m, with m not divisible by p. Then every Sylow p-subgroup has order p raised to a, they are all conjugate, and their number n satisfies two rules: n divides m and n leaves remainder 1 upon division by p.
You try this on orders you meet often. For order 12, which is 4 times 3, n for p equal to 3 divides 4 and is 1 mod 3, so it is 1 or 4. For order 33, n for p equal to 3 divides 11 and is 1 mod 3, so it is 1. For order 20, n for p equal to 5 divides 4 and is 1 mod 5, so it is 1. For order 18, the Sylow 3-subgroup has order 9, the largest power of 3 dividing 18.
You get normality when the count is forced to 1, since a unique Sylow is fixed by conjugation. If all Sylow subgroups are normal and the order is composite, the group is not simple. For order 12, count 1 gives a normal Sylow 3, while count 4 gives eight elements of order 3, leaving a normal Sylow 2. For order 30, Sylow 5 count is 1 or 6, and count 6 leaves 24 elements of order 5, forcing a normal Sylow 3.
You finish small orders the same way each time. Factor the order, list the allowed counts for each prime, and look for a forced 1. For order 15, the Sylow 3 count divides 5 and is 1 mod 3, so it is 1, and the Sylow 5 count divides 3 and is 1 mod 5, so it is 1. Both are normal, the group is a direct product of cyclic groups of coprime order, hence cyclic. For order 6, this method leaves exactly two groups: the cyclic group of order 6 and the permutations of three letters.
You count Sylow subgroups with two arithmetic rules, and a forced unique one gives normality and controls small groups.
2 · Watch
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8 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.