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# 数学代写|抽象代数代写Abstract Algebra代考|MATH4200 The dihedral group $D_3$

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## 数学代写|抽象代数代写Abstract Algebra代考|The dihedral group $D_3$

The dihedral group $D_3$ is the group of symmetries of an equilateral triangle, mentioned in Section 1.3 and discussed in Section 5.7. The group operation is composition-symmetries are combined by performing one then another.

Now, $D_3$ has six elements. Is it isomorphic-structurally identical-to $\mathbb{Z}_6$ ? If you trust intuition based on group tables, you might feel that it is not, because the structures are different: the table for $\mathbb{Z}_6$ has diagonal ‘stripes’, whereas this table has ‘blocks’. But could reordering the elements reveal hidden stripes? One way to answer is to consider group properties. Every cyclic group is commutative. And $D_3$ is not: for instance, $\rho r_1 \neq r_1 \rho$. Reordering cannot make non-commutativity ‘go away’ because it does not change the result of any composition. So $\mathrm{D}_3$ is not cyclic.

Another approach is to consider the definition, which says that a cyclic group is generated by a single element. In $D_3$, the identity $e$ generates the single-element set ${e}$. The rotations $\rho$ and $\rho^2$ each generate the subgroup $\left{e, \rho, \rho^2\right}$. What does a reflection generate? Repeatedly performing any reflection gives the reflection then the identity then the reflection then the identity, and so on. So each reflection generates a two-element subgroup. Thus $D_3$ is not generated by a single element, so is not cyclic.

That raises a question, however. If a single element is not enough to generate $D_3$, how many are needed? What do you think? Can two elements be combined in different ways to give every group element? Does it matter which two? I recommend that before reading on, you make a triangle and experiment.

## 数学代写|抽象代数代写Abstract Algebra代考|More symmetry groups

The obvious group to investigate next is $D_4$, the group of symmetries of a square.

Does the information about $D_3$ generalize to $D_4$ ? Maybe, because $D_3$ and $D_4$ are similarly constructed. But $D_4$ is bigger, so there is ‘room’ for more internal structure. What do you think? What are the subgroups of $D_4$, and are they all cyclic? Can two elements generate $\mathrm{D}_4$ ?

Like $D_3$, the group $D_4$ has a two-element subgroup generated by each reflection and a subgroup $\left{e, \rho, \rho^2, \rho^3\right}$ comprising the rotations and the identity. This time, the rotation subgroup itself has a nontrivial proper subgroup $\left{e, \rho^2\right}$. Can you see this in the table and understand it by imagining transformations?

The group $D_4$ also has a four-element subgroup $\left{e, \rho^2, r_3, r_4\right}$, highlighted below in the table and a mini-table. Is this subgroup cyclic? No: each non-identity element is self-inverse so it cannot have a single generator.

Thus $D_4$ is unlike $D_3$ in having proper subgroups that are not cyclic. The full set of subgroups of $D_4$ is represented below.

Now, $r_3$ and $r_4$ ‘go together’ in that they are diagonal reflections that appear in $\left{e, \rho^2, r_3, r_4\right}$. But their inelegant positioning in the main table suggests that the elements of $D_4$ could be labelled in a more structurally natural way. As for $D_3$, a single rotation and reflection generate $D_4$.

## 数学代写|抽象代数代写Abstract Algebra代考|More symmetry groups

\left 缺少或无法识别的分隔符 包括旋转和身份。这一次，旋转子群本身有一个 非平凡的真子群 $\backslash$ left 缺少或无法识别的分隔符 你能在表中看到这一点并通过想 象转换来理解它吗?

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