Counting prime power subgroups

Mathematics · Abstract Algebra · ages 22-23

Name ______________________ Date ____________

Group order 33. How many Sylow 3-subgroups must there be?
Answer: ______________
Group order 12. How many Sylow 3-subgroups are possible?
- 2 or 3
- 1 or 2
- 3 or 6
- 1 or 4
A group has order 12. Which pair lists the only possible numbers of Sylow 3-subgroups?
- 2 or 3
- 3 or 6
- 1 or 4
If every Sylow subgroup of a finite group of composite order is normal, then the group is not simple.
Circle one: True False
What must be true of any group of order 15?
- It has normal Sylow 3 and Sylow 5 subgroups, so it is cyclic
- It must be the permutations of three letters
- It must be simple
Sylow p-subgroup has order equal to the highest power of p dividing the group order. Is this definition correct?
Circle one: True False
Why is no group of order 12 simple?
- All groups of order 12 are abelian
- Sylow 3 count is 1 or 4; either a normal Sylow exists or counting forces a normal Sylow 2
- Groups of order 12 have no subgroups
- Sylow theorems do not apply
Why is no group of order 12 simple?
- All groups of order 12 are abelian
- The Sylow 3 count is 1 or 4, and both cases force a normal Sylow
- Sylow theorems do not apply to order 12
Which pair is the full list of groups of order 6, up to isomorphism?
- Cyclic of order 6 only
- Cyclic of order 6 and permutations of three letters
- Permutations of three letters and quaternions
Group order 20. How many Sylow 5-subgroups must there be?
Answer: ______________

Answer key

For grown-ups. Fold this page away before handing over the rest.

1 · Count divides 11 and is 1 modulo 3. Divisors are 1 and 11, and 11 leaves 2, so only 1 remains.
1 or 4 · 1 or 4 is correct since the count divides 4 and is 1 modulo 3, leaving only 1 and 4.
1 or 4 · The count divides 4 and is 1 mod 3, which leaves only 1 and 4.
True · A normal Sylow subgroup is proper and nontrivial, which blocks simplicity.
It has normal Sylow 3 and Sylow 5 subgroups, so it is cyclic · Both Sylow counts are forced to 1, so the group is a direct product of cyclic groups of coprime order.
True · This is the defining order condition for Sylow subgroups.
Sylow 3 count is 1 or 4; either a normal Sylow exists or counting forces a normal Sylow 2 · Sylow 3 count is 1 or 4; either a normal Sylow exists or counting forces a normal Sylow 2 is correct, while the other claims are false.
The Sylow 3 count is 1 or 4, and both cases force a normal Sylow · A count of 1 gives a normal Sylow 3, and a count of 4 leaves a normal Sylow 2.
Cyclic of order 6 and permutations of three letters · Every group of order 6 is one of these two, and both occur.
1 · Count divides 4 and is 1 modulo 5. Divisors 1, 2, 4 leave only 1 valid.