Category: Abstract Algebra

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202004240630 Homework 1 (Q2)

Call G a group and e the identity of G. If the order of G is 2, then a^2=e for all a\in G and G is abelian (Prove it yourself). Show that

G is abelian if a^2=e for all a\in G.

Give an example of such a group whose order is greater than 2.

Solution.

Recall that a group G is said to be abelian if xy=yx for all x,y\in G.

For any a,b\in G, there is ab=(bb^{-1})ab(a^{-1}a), because a group satisfies, first, the inverse axiom xx^{-1}=x^{-1}x=e under the notation x^{-1} being the inverse of x, and secondly, the identity axiom ex=xe=x.

Thence by (associativity axiom) x*(y*z)=(x*y)*z I may change brackets and obtain as follows:

\begin{aligned} ab & =(bb^{-1})ab(a^{-1}a) \\ & =b(b^{-1}a)(ba^{-1})a \\ & =b(b^{-1}a)(b^{-1}a)^{-1}a \end{aligned}

Provided that a^2=e for all a\in G, one may deduce a=a^{-1}, b=b^{-1}, etc. Thus,

ab=b(b^{-1}a)(b^{-1}a)a

In addition (b^{-1}a)(b^{-1}a)=(b^{-1}a)^2=e, hence ab=ba, and G abelian.

The product group \mathbb{Z}_2\times \mathbb{Z}_2 is an example of one abelian group whose order is greater than 2 and the order of whose elements is 2.

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202004240606 Homework 1 (Q1)

Let G=\{ a,b,c,d\} and the Cayley table of G be

\begin{aligned} * \quad | &\textrm{} \quad a &\textrm{}\quad b \quad &\textrm{}\quad c &\textrm{}\quad d \\ \hline a \quad | &\quad a &\quad b \quad &\quad c &\quad d \\ b \quad | &\quad b &\quad a \quad &\quad c &\quad d \\ c \quad | &\quad c &\quad b \quad &\quad a &\quad d \\ d \quad | &\quad d &\quad d \quad &\quad b &\quad c \end{aligned}

Is (G,*) a group?

Attempts.

I recall that a group (G,*) is a set G altogether with an operation *:G\times G\rightarrow G, (x,y)\mapsto x*y such that the following axioms be satisfied:

i. (Associativity) (x*y)*z=x*(y*z);

ii. (Identity) There exists e\in G such that e*x=x*e=x for all x\in G; and

iii. (Inverse) For each a\in G, there exists b\in G such that a*b=b*a=e.

Let me check them one by one.

i. For x,y,z\in \{ a,b,c,d \}, it suffices to check 4\times 4\times 4=64 operations. And I found that the following operation failed axiom (i):

d*(c*b)=d*b=d but (d*c)*b=b*b=a.

That said, G under the operation * is not a group.

ii. By inspection, the identity of G exists, and that is a because a*x=x*a=x for all x\in \{ a,b,c,d\}. Axiom (ii) is thus satisfied.

iii. There exists d\in G, such that for all x^i\in\{ a,b,c,d\}=G,

d*x^i=x^i*d\neq a.

The claim above is validated by simply checking the identity found to be a in part ii. cannot be found in the entries, and by the fact that a is unique.

Axioms i. and iii. having been failed, I may conclude that G under the operation * cannot form a group.