Prove that an Abelian group of order 33 is cyclic. Does your proof hold when 33 is replaced by p

step 1 group A such that all elements other than the identity are order 2. We want to first show that A

Prove that an Abelian group of order 33 is cyclic. Does your...

Key Concepts

Finite Abelian Group Structure

A finite Abelian group can be expressed as a direct sum of cyclic groups whose orders are prime powers. In the specific case where the group order is the product of two distinct primes, this decomposition naturally leads to a structure that is isomorphic to the direct product of two cyclic groups, thereby simplifying the group’s structure considerably.

Unique Sylow Subgroups

When a group’s order is the product of two distinct primes, each corresponding Sylow subgroup must have a unique instance due to Sylow’s theorems. The uniqueness implies that these subgroups are normal. In an Abelian context, this normality is inherent, helping further to establish that the group can be written as the product of its Sylow subgroups.

Direct Product of Cyclic Groups

The direct product of two cyclic groups whose orders are coprime is itself cyclic. This important fact, often justified using the Chinese Remainder Theorem, shows that the group is isomorphic to a cyclic group whose order is the product of the relatively prime orders, thereby proving the group is cyclic.

Coprimality

The concept of coprimality is central in this proof because when the orders of the constituent cyclic groups (derived from the prime factors of the group order) are coprime, their direct product has a structure that is cyclic. This property is not unique to a specific example but applies generally to products of groups of orders that are pairwise relatively prime.

Transcript

00:01 Group a such that all elements other than the identity are order 2.

00:04 We want to first show that a must be a billion, and second show that it must have this subgroup where x and y are distinct elements of a.

00:15 Then lastly, third, actually, we want to show that if we have some group that has order 2 times p, where p is a prime greater than 2, then there must exist an element in b such that the order is p and not just two.

00:35 So let's first look at whether or not a billion.

00:42 Let x and y be elements of a such that x times y equals z.

00:54 Then let's let's left multiply by by x so then we get that y equals x x times z but then we can also write multiply the same expression by y then we get x must equal z times y let's then consider then that y times x would then equal x times z times z times y where here of course we're still assuming x y and z none of them equal the identity so then we have x times z squared times y which equals x times y and we've just shown commutativity, so thus a is a beelian.

02:06 If we have two distinct elements, so let x and y be in a, such that x does not equal y, does not equal the identity, so they're distinct elements.

02:25 We can then form from this so x is its own inverse y is its own inverse and then x times y with this we will have this does form a subgroup of a since it contains the identity x is its own inverse y is its own inverse x times y is its own inverse so this is closed so, yes, we will have those subgroups for each subset or each two distinct elements of a.

03:09 Now let's look at this last part...

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