/sci/ - Science & Math

[Anonymous No.4488924]

Hey,

You guys might remember me from earlier. I'm studying for a group theory test in a few days by proving all the theorems in my book I don't know proofs for. I need help with one more...

If every group of order n > 1 is cyclic, then n is either prime or a product of
distinct primes.


Thanks!

[Anonymous No.4488926]

bump

[Anonymous No.4488940]

bump, I got a lot of great help before....hoping someone can give me a hand. Proving n can be prime is fairly trivial as all groups of prime order are cyclic, I'm just having trouble with the "product of distinct primes" possibility.

[Anonymous No.4488949]

Note that in, for example, <span class="math">\mathbb{Z}/6\mathbb{Z} \cong \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}[/spoiler]. Specifically, the latter has an element <span class="math">1 \times 1[/spoiler] of order 6. In general, the direct sum of groups has an element with order equal to the least common multiple of the orders of the summands.

[Anonymous No.4488962]

>>4488949
Now, what you need to prove is that for orders with a factor <span class="math">p^2[/spoiler], then there is a group with that order that is not cyclic. You should be convinced that in the direct product of cyclic groups, any element's order divides the lcm of the orders of the summands. So, let <span class="math">p^n[/spoiler] be the highest power of p dividing |G|. Then if we let G be the abelian group<span class="math">\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z} \times \cdots[/spoiler] then it is easy to see that the least common multiple of the summands is at most |G|/p, and therefore G has no element of order p.

[Anonymous No.4488964]

>>4488962
sorry, the end of this is: G has no element of order |G|

[Anonymous No.4488969]

>>4488964
Alternatively, you can construct the elementary abelian group (or some other group containing <span class="math">Z/pZ \times Z/pZ[/spoiler] as a subgroup). Since that direct product is not cyclic, G cannot be cyclic, because subgroups of cyclic groups are cyclic.

[Anonymous No.4488983]

whoa, I'm studying for a group theory test in a few days too. Mine's on isomorphisms, direct products, factor groups, and homorphisms...you?

shit's crazy I've been studying like mad and I think it's finally starting to come together. Fuck those problems where two groups aren't isomorphic because they don't have the same number of elements of the same order.


anyone know anything online with good practice problems like these or anything relevant?

[Anonymous No.4488986]

>>4488983
Memorize the symmetric groups and all of their subgroups for n < 20

[Anonymous No.4488996]

>>4488986
lol

[Anonymous No.4489143]

http://en.wikipedia.org/wiki/Fundamental_theorem_of_cyclic_groups