Math 4441/6441 (Spring 2008) Homework Set #7 (Due 03/13) Hints
Problem 1 (10 points).Let G6=∅be a set with a (binary) operation ∗that is associative.
(Thus a∗b∈Gand (a∗b)∗c=a∗(b∗c) for all a, b, c ∈G.) Further assume
(I) There exists a (special) element e∈Gsuch that a∗e=afor all a∈G.
(II) For every a∈G, there exists a0∈Gsuch that a∗a0=e.
Show Gis a group under the operation ∗. Show your work in detail.
Hint. It suffices to show e∗a=aand a0∗a=efor all a∈G, where eand a0are the same
as described in (I) and (II) above.
You may mimic the proofs of Lemma 2.44 and Lemma 2.45(i)(ii) on page 127 to show
e∗a=aand a0∗a=efor all a∈G. (It won’t work if you simply copy the proofs of
Lemma 2.44 and Lemma 2.45(i)(ii) verbatim. The reason is that the assumption and
conclusion in this problem differ from those in Lemma 2.44 and Lemma 2.45(i)(ii).)
To be specific, first show that if an element x∈Gsatisfies x∗x=x, then x=e(cf.
Lemma 2.44). Then use it to prove a0∗a=efor all a∈G(cf. Lemma 2.45(i)). Finally,
show e∗a=afor all a∈G(cf. Lemma 2.45(ii)). Other approaches are welcome.
Problem 2 (10 points).Let G6=∅be a set with a (binary) operation ∗that is associative.
(Thus a∗b∈Gand (a∗b)∗c=a∗(b∗c) for all a, b, c ∈G.) Further assume
(I) For every a, b ∈G, there exists x∈Gsuch that a∗x=b.
(II) For every a, b ∈G, there exists y∈Gsuch that y∗a=b.
Show Gis a group under the operation ∗. Show your work in detail.
Hint. We need to do the following: (1) find a (special) element e∈Gsuch that e∗a=afor
all a∈G; and (2) for every a∈G, show there exists an element a0∈Gsuch that a0∗a=e.
To accomplish (1), choose an element u∈G(and then fix it). Then by (II), there exists
an element, which we denote by e, such that e∗u=u. Then you need to show that this e
satisfies e∗a=afor all a∈Gby applying (I) or (II) (which one?) to elements uand a.
To show (2), study (I) and (II) carefully. Apply one of them (which one?).
Problem 3 (6441 problem, 5 points).Let Gbe a group (under an operation ∗). (For
a, b ∈G, we may write a∗b=ab for short.) Assume (ab)3=a3b3and (ab)5=a5b5for all
a, b ∈G. Show Gis abelian.
Hint. Let a, b ∈Gbe two arbitrary elements. It suffices to show ab =ba.
The equation (ab)3=a3b3expands to ababab =aaabbb, i.e., a(baba)b=a(aabb)b. By
cancellation, what can you get? That is, baba =?
Similarly, the equation (ab)5=a5b5expands to ababababab =aaaaabbbbb. By cancellation,
what can you get? That is, babababa =?
Now, what is the relation between babababa and baba? What will this relation translate
to in light of what you have got above? “Play” with the derived equations (according to the
rules) and be ready to use the rule of cancellation over and over again (if applicable). You
goal is to show ab =ba (according to the rules).
Other approaches are available and welcome.
For solutions, click here.
Notice. All quoted results, such as Theorem x.y, are from the textbook A First Course in
Abstract Algebra with Applications (3rd Ed.) by J. J. Rotman, Pearson Prentice Hall.
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