Day: May 10, 2023

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202305101606 Problem 3.16.1

Let G=\{ \mathbf{e},a,a^2,a^3\} be a cyclic group of order 4 and G' be the multiplicative group \{1,-1,\mathbf{i},-\mathbf{i}\} of 4 complex numbers.

i. Tabulate two isomorphisms from G to G'.
ii. Explain why there are no other isomorphisms from G to G'.

Extracted from A. P. Hillman. (1999). Abstract Algebra A First Undergraduate Course.

Background.

The group G is cyclic if and only if every element of G can be expressed as the power of one element of G:

\exists\, g\in G,\,\forall\, h\in G:\enspace h=g^n\textrm{ for some }n\in\mathbb{Z}.

if and only if it is generated by one element g\in G called a generator of G:

G=\langle g\rangle.

ProofWiki on Cyclic Group

Let (F,+,\times ) be a field and let F^*:=F\backslash\{ 0\} be the set F less its zero. The group (F^*,\times ) is known as the multiplicative group of F.

ProofWiki on Multiplicative Group

A bijective homomorphism is called an isomorphism; that is, a group isomorphism \theta from G to G' is a bijection \theta such that \theta (ab)=\theta (a)\theta (b) for all a and b in G.

Text on 3.2 Group Isomorphism, pg.141

Let \theta be a group isomorphism from G to G'. Then:

(a) \mathbf{e}\stackrel{\theta}{\mapsto}\mathbf{e}', where \mathbf{e} and \mathbf{e}' are the identities of G and G'.
(b) If a\stackrel{\theta}{\mapsto}a', then a^n\stackrel{\theta}{\mapsto}(a')^n for all integers n. In particular,

a^{-1}\stackrel{\theta}{\mapsto}(a')^{-1}.

(c) If a\stackrel{\theta}{\mapsto}a', then a and a' have equal orders.

Text on 3.2 Group Isomorphism, pg.143

Roughwork.

\begin{array}{c|cccc} (G,*) & \mathbf{e} & a & a^2 & a^3 \\\hline \mathbf{e} & \mathbf{e} & a & a^2 & a^3 \\ a & a & a^2 & a^3 & \mathbf{e} \\ a^2 & a^2 & a^3 & \mathbf{e} & a \\ a^3 & a^3 & \mathbf{e} & a & a^2 \\ \end{array}

\begin{array}{c|cccc} (G',\times ) & 1 & -1 & \mathbf{i} & -\mathbf{i} \\\hline 1 & 1 & -1 & \mathbf{i} & -\mathbf{i} \\ -1 & -1 & 1 & -\mathbf{i} & \mathbf{i} \\ \mathbf{i} & \mathbf{i} & -\mathbf{i} & -1 & 1 \\ -\mathbf{i} &-\mathbf{i} & \mathbf{i}} & 1 & -1 \\ \end{array}

Observe that

G=\langle a\rangle; G'=\langle \mathbf{i}\rangle=\langle -\mathbf{i}\rangle.

This problem is not to be attempted.