anon anon
2022-06-13
Ian Adams
Skilled2022-06-14Added 163 answers
Step 1
Solution: If G has no maximal subgroups, I(G) is taken to be G.⇒I(G) characterizes G in this case. Assume G has at least one maximal subgroup MLet α∈sub(G), Then α(M) is a maximal subgroup of G.If N≤G with α(M)≤N≤G, Then M≤α-1(N)≤G⇒α-1(N)=M or α-1(N)=G⇒N=α(M) or N=GSo, α(M) is maximal.
Step 2
1)α-1∈Sub(G), so α-1(M) is maximal.2)(α.α-1)(M)=(α-1.α)(M)So, α permutes the maximal subgroups of G.Let x∈I(G), then x∈∩M≤GMSo, α(x)∈α(∩M≤GM)=∩M≤Gα(M)=∩M≤GM=I(G)
How to find out the mirror image of a point?
Generators of a free group
If G is a free group generated by n elements, is it possible to find an isomorphism of G with a free group generated by n-1 (or any fewer number) of elements?
How many 3/4 Are in 1
Convert 10 meters to feet. Round your answer to the nearest tenth
6. Reduce the following matrix to reduced row echelon form:
Let v be a vector over a field F with zero vector 0 and let s,T be a substance of V .then which of the following statements are false
Describe Aut(Zp), the automorphism group of the cyclic group Zp where p is prime. In particular find the order of this group. (Hint: A generator must map to another generator)
Let a and b belong to a ring R and let m be an integer. Prove that m(ab) = (ma)b = a(mb)