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State and prove Lagrange's theorem?
THEOREM:The order of a subgroup H of group G divides the order of G.First we need to define the order of a group or subgroupDefinition:If G is a finite group (or subgroup) then the order of G is the number of elements of G.Lagrange's Theorem simply states that the number of elements in any subgroup of a finite group must divide evenly into the number of elements in the group. Note that the {A, B} subgroup of the Atayun-HOOT! group has 2 elements while the Atayun-HOOT! group has 4 members. Also we can recall that the subgroups of S3, the permutation group on 3 objects, that we found cosets of in the previous chapter had either 2 or 3 elements -- 2 and 3 divide evenly into 6.A consequence of Lagrange's Theorem would be, for example, that a group with 45 elements couldn't have a subgroup of 8 elements since 8 does not divide 45. It could have subgroups with 3, 5, 9, or 15 elements since these numbers are all divisors of 45.Now that we know what Lagrange's Theorem says let's prove it. We'll prove it by extablishing that the cosets of a subgroup aredisjoint -- different cosets have no member in common, andeach have the same number of members as the subgroup.This leads to the conclusion that a subset with n elements has k cosets each with n elements. These cosets do not overlap and together they contain every element in the group. This means that the group must have kn elements -- a multiple of n. We'll accomplish our proof with two lemmas. Lemma:If H is a finite subgroup of a group G and H contains n elements then any right coset of H contains n elements.Proof:For any element x of G, Hx = {h • x | h is in H} defines a right coset of H. By the cancellation law each h in H will give a different product when multiplied on the left onto x. Thus each element of H will create a corresponding unique element of Hx. Thus Hx will have the same number of elements as H.Lemma:Two right cosets of a subgroup H of a group G are either identical or disjoint.Proof:Suppose Hx and Hy have an element in common. Then for some elements h1 and h2 of Hh1 • x = h2 • yThis implies that x = h1-1 • h2 • y. Since H is closed this means there is some element h3 (which equals h1-1 • h2) of H such that x = h3 • y. This means that every element of Hx can be written as an element of Hyby the correspondenceh • x = (h • h3) • yfor every h in H. We have shown that if Hx and Hy have a single element in common then every element of Hx is in Hy. By a symmetrical argument it follows that every element of Hy is in Hx and therefore the "two" cosets must be the same coset.Since every element g of G is in some coset (namely it's in Hg since e, the identity element is in H) the elements of G can be distributed among H and its right cosets without duplication. If k is the number of right cosets and n is the number of elements in each coset then |G| = kn.Alternate Proof:In the last chapter we showed that a • b-1 being an element of H was equivalent to a and b being in the same right coset of H. We can use this Idea establish Lagrange's Theorem.Define a relation on G with a ~ b if and only if a • b-1 is in H. Lemma: The relation a ~ b is an equivalence relation.Proof:We need to establish the three properties of an equivalence relation -- reflexive, symmetrical and transitive.(1) Reflexive:Since a • a-1 = e and e is in H it follows that for any a in Ga ~ a(2) Symmetrical:If a ~ b then a • b-1 is in H. Then the inverse of a • b-1 is in H. But the inverse of a • b-1 is b • a-1 sob ~ a(3) Transitive:If a ~ b and b ~ c then both a • b-1 and b • c-1 are in H. Therefore their product (a • b-1) • (b • c-1) is in H. But the product is simply a • c-1. Thusa ~ cAnd we have shown that the relation is an equivalence relation.It remains to show that the (disjoint) equivalence classes each have as many elements as H.Lemma:The number of elements in each equivalence class is the same as the number of elements in H.Proof:For any a in G the elements of the equivalence class containing a are exactly the solutions of the equationa • x-1 = hWhere h is any element of H. By the cancellation law each member h of H will give a different solution. Thus the equivalence classes have the same number of elements as H.One of the imediate results of Lagrange's Theorem is that a group with a prime number of members has no nontrivial subgroups. (why?)Definition:if H is a subgroup of G then the number of left cosets of H is called the index of H in G and is symbolized by (G:H). From our development of Lagrange's theorem we know that|G| = |H| (G:H)
Prove that, 36^7−6^13 is divisible by 30.?
help me
How do you prove 5 over 6 is equal to 10 over 12?
Divide each fraction. 5/6 = 0.833333... and 10/12 = 0.833333... .
Does 6 dollar click pays?
i m not confirm yet...bt i have an idea to prove it....normally if we work in an account we have to work at least 6 month regularly to make 6000 dollar as we got daily 36 dollar..but with the help of reference we can do it only in 6 days(is'nt it amazing?)...but u have to work hard these 6 days & then check does it pay or not...u need not wait 6 months to prove does it pay? any Q?
What is the process of atherosclerosis in order of 1 - 6?
-5
Can you prove that these numberc can be written in arbitrary order?
The 10 digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 are written in an arbitrary order. Prove that one can always remove 6 digits such that the remaining 4 digits are ordered monotonically. Help...
If d divides the order of a group does the group has a subgroup of order d?
The general answer is no. Consider A4={(1),(12)(34),(13)(24),(14)(23),(123),(124),(132),(134),(142),(143),(234),(243)}. The subgroups of A4 are: A4, , , , =, =, =, =, {(1),(12)(34),(13)(24),(14)(23)}, {(1)}. The order of A4 is 12, the order of , and is 2, the order of =, =, = and = is 3, the order of {(1),(12)(34),(13)(24),(14)(23)} is 4, and the order of is 1. Clearly there are no subgroups of order 6, but 6 definitely divides the order of A4. The statement is true for all finite abelian groups, and when d is a power of a prime (i.e., when d=pk for a prime p and a non-negative integer k).
Remainder of 4n where n is any natural number when divided by 6 is always 4. Prove this?
It is REM(4n/6)=4. Prove. Correction.
Which group is chromium in?
Chromium is in the group 6 of the periodic table of Mendeleev.
What are the release dates for Chopped - 2007 Prove It on the Plate 6-2?
Chopped - 2007 Prove It on the Plate 6-2 was released on: USA: 11 January 2011
What is the group is oxygen in?
Group 6
What is the group name of the elements in group 6?
Chromium Group
What group is the element oxygen in?
Group 6
State and prove Lagrange's theorem?
THEOREM:The order of a subgroup H of group G divides the order of G.First we need to define the order of a group or subgroupDefinition:If G is a finite group (or subgroup) then the order of G is the number of elements of G.Lagrange's Theorem simply states that the number of elements in any subgroup of a finite group must divide evenly into the number of elements in the group. Note that the {A, B} subgroup of the Atayun-HOOT! group has 2 elements while the Atayun-HOOT! group has 4 members. Also we can recall that the subgroups of S3, the permutation group on 3 objects, that we found cosets of in the previous chapter had either 2 or 3 elements -- 2 and 3 divide evenly into 6.A consequence of Lagrange's Theorem would be, for example, that a group with 45 elements couldn't have a subgroup of 8 elements since 8 does not divide 45. It could have subgroups with 3, 5, 9, or 15 elements since these numbers are all divisors of 45.Now that we know what Lagrange's Theorem says let's prove it. We'll prove it by extablishing that the cosets of a subgroup aredisjoint -- different cosets have no member in common, andeach have the same number of members as the subgroup.This leads to the conclusion that a subset with n elements has k cosets each with n elements. These cosets do not overlap and together they contain every element in the group. This means that the group must have kn elements -- a multiple of n. We'll accomplish our proof with two lemmas. Lemma:If H is a finite subgroup of a group G and H contains n elements then any right coset of H contains n elements.Proof:For any element x of G, Hx = {h • x | h is in H} defines a right coset of H. By the cancellation law each h in H will give a different product when multiplied on the left onto x. Thus each element of H will create a corresponding unique element of Hx. Thus Hx will have the same number of elements as H.Lemma:Two right cosets of a subgroup H of a group G are either identical or disjoint.Proof:Suppose Hx and Hy have an element in common. Then for some elements h1 and h2 of Hh1 • x = h2 • yThis implies that x = h1-1 • h2 • y. Since H is closed this means there is some element h3 (which equals h1-1 • h2) of H such that x = h3 • y. This means that every element of Hx can be written as an element of Hyby the correspondenceh • x = (h • h3) • yfor every h in H. We have shown that if Hx and Hy have a single element in common then every element of Hx is in Hy. By a symmetrical argument it follows that every element of Hy is in Hx and therefore the "two" cosets must be the same coset.Since every element g of G is in some coset (namely it's in Hg since e, the identity element is in H) the elements of G can be distributed among H and its right cosets without duplication. If k is the number of right cosets and n is the number of elements in each coset then |G| = kn.Alternate Proof:In the last chapter we showed that a • b-1 being an element of H was equivalent to a and b being in the same right coset of H. We can use this Idea establish Lagrange's Theorem.Define a relation on G with a ~ b if and only if a • b-1 is in H. Lemma: The relation a ~ b is an equivalence relation.Proof:We need to establish the three properties of an equivalence relation -- reflexive, symmetrical and transitive.(1) Reflexive:Since a • a-1 = e and e is in H it follows that for any a in Ga ~ a(2) Symmetrical:If a ~ b then a • b-1 is in H. Then the inverse of a • b-1 is in H. But the inverse of a • b-1 is b • a-1 sob ~ a(3) Transitive:If a ~ b and b ~ c then both a • b-1 and b • c-1 are in H. Therefore their product (a • b-1) • (b • c-1) is in H. But the product is simply a • c-1. Thusa ~ cAnd we have shown that the relation is an equivalence relation.It remains to show that the (disjoint) equivalence classes each have as many elements as H.Lemma:The number of elements in each equivalence class is the same as the number of elements in H.Proof:For any a in G the elements of the equivalence class containing a are exactly the solutions of the equationa • x-1 = hWhere h is any element of H. By the cancellation law each member h of H will give a different solution. Thus the equivalence classes have the same number of elements as H.One of the imediate results of Lagrange's Theorem is that a group with a prime number of members has no nontrivial subgroups. (why?)Definition:if H is a subgroup of G then the number of left cosets of H is called the index of H in G and is symbolized by (G:H). From our development of Lagrange's theorem we know that|G| = |H| (G:H)
Which elements in period 3 are most likely to form negative ions?
Chlorine fist as it is in group 7. Then group 6 to 1 in that order. i.e. Sulphur, Phosphorus, Silicon, Aluminium, Magnesium, then sodium.
What group made up of six nonmetals?
Group 6
What group made up six nonmetals?
Group 6
