Viewing ⟨3⟩and ⟨12⟩as subgroups of Z, prove that ⟨3⟩/⟨12⟩is isomorphic to Z4 . Similarly, prove

Viewing ⟨3⟩and ⟨12⟩as subgroups of Z, prove that ⟨3⟩/⟨12⟩is...

Key Concepts

Group Isomorphism

A group isomorphism is a bijective map between two groups that preserves the group operation. Demonstrating that a quotient group is isomorphic to a known cyclic group, such as Z? or Z?, implies that they share the same algebraic structure. This concept is essential when generalizing results to show that ?k?/?nk? is isomorphic to Z?.

Generalization to Arbitrary Integers

The generalization involves recognizing that for any integers k and n, the subgroup ?nk? is a subgroup of ?k? with an index equal to n. This implies that the quotient group ?k?/?nk? is cyclic of order n, and thus is isomorphic to Z?, establishing a broad principle within the structure theory of cyclic groups.

Cosets

Cosets are the sets formed by adding a fixed element from the group to every element of a subgroup. They play a critical role in defining quotient groups by partitioning the group. Understanding cosets allows one to compute the index of the subgroup and helps in establishing isomorphisms by revealing the cyclic nature of the quotient.

Subgroups of Integers

Subgroups of Z, such as those generated by an integer k (denoted ?k?), consist of all integer multiples of k. The fact that every subgroup of Z has the form kZ is a consequence of the structure theorem for cyclic groups and greatly simplifies the understanding of their properties and relationships.

Cyclic Groups

Cyclic groups are groups that can be generated by a single element. In the context of the integers, the group Z itself is cyclic, and every subgroup of Z is also cyclic. This facilitates the analysis of subgroups like ?3? or ?12? in terms of their generators and gives rise to structurally simple factor groups.

Quotient Groups

A quotient group is formed by partitioning a group into cosets relative to a normal subgroup. In Z, every subgroup is normal, making the quotient group well-defined. The quotient ?k?/?nk?, for example, is constructed by considering cosets of ?nk? in ?k?, and its structure is captured by the number of distinct cosets, which is tied to the index of ?nk? in ?k?.

Transcript

00:01 For this question, we want to look at the quaternion group, which is sometimes written as q8.

00:09 And we want to determine subgroups, the co -sets, and look at whether or not it's isomorphic to d4.

00:15 The first step in this process is to write out the multiplication table.

00:23 You can see that it's not a billion just by the setup that ij is going to be equal to k, but j times i is negative k so if we write out the multiplication table we can identify some subgroups remember this is going to be by looking at like subsets or rows and columns where the products are all contained within like the same set so for instance we can look at this subgroup here and we see we don't have any jays or ks that show up within the subgroup thus that is in fact a subgroup.

01:05 We have four subgroups within the quaternian group.

01:10 We have this subgroup of order 2.

01:14 And then we have these three subgroups of order 4.

01:20 So for this first one, let's call it h .i.

01:23 We could write out the cosets.

01:25 They look like this.

01:26 And they do, in fact, cover all elements of the group.

01:32 This subgroup is normal.

01:34 We could do that by if we look at for any element, a in q, if we consider a, h1, sorry, h1 times a inverse, which here we could do this for any ijk or a negative 1 or negative 1, and you're going to get that this equals h -i back in the end.

02:04 Because here the inverses, of course, are like i -inverse is negative -vi, etc.

02:13 So this one is normal, and it is a typo in the question as well, because these subgroups of order 4 are also normal subgroups.

02:25 And this is like the fun feature of the quaternian group that all four subgroups are in fact normal.

02:32 We could do that by say let's look at h .i.

02:36 It's pretty obvious, just as with h1, that if we consider the product of any other element with the 1 and negative 1, we'll get 1 and negative 1 back.

02:47 But we could look at, let's say, i times, let's do negative j...

Please give Ace some feedback