Assignment 3
( \newcommand{\kernel}{\mathrm{null}\,}\)
For the following, either show that G is a group with a given operation or list the condition(s) that fail.
- G=R with the binary operation a⋆b=a+b−ab for a,b∈R.
- G=3Z={3m|m∈Z} with addition.
- G={4,8,12,16} with multiplication mod(20).
- Answer
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1. LetG:=R, and let a,b,c∈G. Let an operation ⋆) onG be defined asa⋆b:=a+b−a⋅b. \\
Is (G,⋆) closed? (⋆) is comprised of (+), (−), and (⋅) over real numbers. These operations are closed over R. Therefore, sinceG=R, (G,⋆) is closed.
Is(G,⋆) associative? Consider the following:a⋆(b⋆c)=a⋆(b+c−bc),=a+(b+c−bc)−a(b+c−bc),=a+b+c−bc−ab−ac+abc,=(a+b−ab)+c−c(a+b−ab),=(a+b−ab)⋆c,=(a⋆b)⋆c.
Therefore, (G,⋆) is associative.
DoesG contain the identity e? Consider 0∈G.
a⋆0=a+0−a⋅0,=a+0−0,=a.0⋆a=0+a−0⋅a,=0+a−0,=a.
Sincea⋆0=0⋆a=a for any a∈G, and 0∈G;e=0 is the identity of (G,⋆).
Does each element of G have an inverse in G? Let's assume for contradiction that a−1∈G exists for all a∈G. Then we can say that:
a⋆a−1=0,a+a−1−a⋅a−1=0,a−1(1−a)=−a,a−1=−a1−a.
Now consider where a=1. In this case, 1−1=11−1=10. Clearly 10∉G, so1−1∉G. This is a contradiction. Therefore, there exists a∈G such that a−1∉G.
Since(G,⋆) is closed, has an identity, is associative, but does not contain an inverse for each element; (G,⋆) is not a group.\ ((G, \star)\) is a monoid.
2. Let G:=3Z={3m:m∈Z}. Let m,n,p∈Z. Clearly 3m,3n,3p∈3Z by definition of 3Z. Consider (G,+).
Is (G,+) closed? Consider the following:
3m+3n=3(m+n).
Since Z is closed under addition, (m+n)∈Z, and 3(m+n)∈G. Therefore, (G,+) is closed.
Is (G,+) associative? Consider the following:
3m+(3n+3p)=3(m+(n+p)),=3((m+n)+p),=(3m+3n)+3p.
Therefore, (G,+) is associative.
DoesG contain the identity e? Consider 0∈G.
3m+0=3m.0+3m=3m.
Clearly 3m+0=0+3m=3m. Therefore, e=0 is the identity of (G,+).
Does(G,+) contain the inverse for each element inG? Consider 3(−n) as the inverse for 3n. Since −n∈Z, 3(−n)∈G.
3n+3(−n)=3(n−n)=3(0)=0=e. and 3(−n)+3n=3((−n)+n)=3(0)=0=e.
Therefore 3n+3(−n)=3(−n)+3n=e. So for any a∈G, there exists a−1∈G.
Since (G,+) is closed, associative, and contains the identity and an inverse for every element, (G,+) is a group.
3. The Cayley table of G with multiplication mod(20):
⋅(mod20) 4 8 12 16 4 16 12 8 4 8 12 4 16 8 12 8 16 4 12 16 4 8 12 16 Is (G,⋅(mod(20)) closed? Yes, we can see from the Cayley table that all results of ⋅(mod(20) are in G.
Does G contain the identity for ⋅(mod(20) ? Again, by the Cayley table, we can see that the identity e=16.
Is (G,⋅(mod(20)) associative? Yes. We know by the properties of Z that multiplication of integers is associative. Also, by the properties of modular arithmetic that the modulo operation has no impact on associativity. Therefore, (G,⋅(mod(20)) is associative. \\
Does G contain an inverse for each element under ⋅(mod(20)? Yes, consider the following:
4⋅4≡16mod(20)⟹4−1=4.8⋅12≡16mod(20)⟹8−1=12.12⋅8≡16mod(20)⟹12−1=8.16⋅16≡16mod(20)⟹16−1=16.
So clearly for each a∈G, there exists a−1∈G. Since (G,⋅(mod(20)) is closed, associative, and contains the identity and an inverse for every element, (G,⋅(mod(20)) is a group.
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Decide whether the following structures form a group or not by looking at the Cayley table.
- {±1,±i} with multiplication, Where i is the complex number such that i2=−1.
- 1=[1001], I=[01−10], J=[0ii0], K=[i00−i], where i is the complex number such that i2=−1.
Define Q8={±1,±I,±J,±K}. Q8 with matrix multiplication.
- Answer
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1. Let G={1,−1, i,−i} with complex multiplication.
We are to determine whether the set G forms a group (G,⋅) under multiplication, where i is the imaginary unit with i2=−1. We will construct and examine the Cayley table to verify the group properties.
Since it is closed, has an identity, and every element has an inverse according to the table, multiplication is associative on 2×2 matrices over the complex numbers; it is a group.
2. Q8={±1,±I,±J,±K}, where 1=[1001], I=[01−10], J=[0ii0], K=[i00−i], where i is the complex number such that i2=−1.
Since it is closed, has an identity, and every element has an inverse according to the table, multiplication is associative on 2×2 matrices over the complex numbers; it is a group.
Prove or disprove: Let H and K be subgroups of the group G. Then
- H∪K is a subgroup of G.
- H∩K is a subgroup of G.
- Answer
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1. No, H∪K is a subgroup of G.
Answers may vary.
Counterexample:
Let G=(Z6,+(mod6)).
Thus eG=0.
We shall show that H∪K is not a subgroup of G.
Consider Z6={0,1,2,3,4,5}.
Let H=({0,2,4},+(mod6)).
Note H≤G sinceeG=0∈H and ∃1h−1,∀h∈H.
Let K=({0,3},+(mod6)).
Note K≤G since eK=0∈H and ∃1k−1,∀k∈K.
Thus H∪K=({0,2,3,4},+(mod6)).
Let a=2 andb=3, where 2,3∈H∪K.
However, since 2+3≡5(mod6)∉H∪K, H∪K is not a group.
Thus, H∪K is not a subgroup ofG.
2. Yes, H∩K is a subgroup of G .
Proof:
Let e be the identity of G .
Then e is the identity of both H and K .
Therefore e∈H∩K .
Let g1,g2∈H∩K .
Then we shall show that g1g−12∈H∩K .
Since g1,g2∈H∩K , g1,g2∈H and g1,g2∈K .
Since H≤G and K≤G , g1,g2,g−11,g−12 are in both H and K .
Hence g1g−12∈H∩K .
Therefore H∩K≤G .◻
1. Show that U(20)≠<k> for any k in U(20). [Hence U(20) is not cyclic.]
2. Prove that if G is an abelian group, then for each element a∈G, C(a)={x∈G|xa=ax} is a subgroup of G.
- Answer
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1.
U(20)={1,3,7,9,11,13,17,19} and |U(20)|=8 .
We shall show thatU(20)≠k,k∈U(20) .
k
Calculations
Order
ofk
1
1≡1(mod20)
1
3
31≡3(mod20) ;32≡9(mod20) ;33≡7(mod20) ;34≡1(mod20)
4
7
71≡7(mod20) ;72≡9(mod20) ;73≡3(mod20) ;74≡1(mod20)
4
9
91≡9(mod20) ;92≡1(mod20)
2
11
111≡11(mod20) ;112≡1(mod20)
2
13
131≡13(mod20) ;132≡9(mod20) ;133≡17(mod20) ;134≡1(mod20)
4
17
171≡17(mod20) ;172≡9(mod20) ;173≡13(mod20) ;174≡1(mod20)
4
19
191≡19(mod20) ;192≡1(mod20)
2
Since |k|≠|U(20)|,∀k∈U(20) , there is no k∈U(20) that generates U(20) .
2. Since G is abelian, for each a∈G, C(a)={x∈G|xa=ax}=G is a subgroup of G.