Question

Show that if G is a nilpotent group and 1 ≠ N, and N is normal in G, then N ∩ Z(G) ≠ 1.

Answer #1

1) Let G be a group and N be a normal subgroup. Show that if G
is cyclic, then G/N is cyclic. Is the converse true?
2) What are the zero divisors of Z6?

Let N be a nilpotent mapping V and letγ:V→V be an isomorphism.
1.Show that N and γ◦N◦γ−1 have the same canonical form 2. If M is
another nilpotent mapping of V such that N and M have the same
canonical form, show that there is an isomorphism γ such that
γ◦N◦γ−1=M

let
H nilpotent normal subgroup of G and f be automorphism map from G
to G ,then f(H) is nilpotent normal subgroup of G

(Linear Algebra)
A n×n-matrix is nilpotent if there is a "r" such that
Ar is the nulmatrix.
1. show an example of a non-trivial, nilpotent 2×2-matrix
2.let A be an invertible n×n-matrix. show that A is not
nilpotent.

Show that if G is a group, H a subgroup of G with |H| = n, and H
is the only subgroup of G of order n, then H is a normal subgroup
of G.
Hint: Show that aHa-1 is a subgroup of G
and is isomorphic to H for every a ∈ G.

Suppose : phi :G -H is a group isomorphism . If N is a normal
subgroup of G then phi(N) is a normal subgroup of H. Prove it is a
subgroup and prove it is normal?

Let G be a finite group, and suppose that H is normal subgroup
of G.
Show that, for every g ∈ G, the order of gH in G/H must divide
the order of g in G.
What is the order of the coset [4]42 +
〈[6]42〉 in Z42/〈[6]42〉?
Find an example to show that the order of gH in G/H does not
always determine the order of g in G. That is, find an example of a
group G, and...

A subgroup H of a group G is called a normal subgroup if gH=Hg
for all g ∈ G. Every Group contains at least two normal subgroups:
the subgroup consisting of the identity element only {e}; and the
entire group G. If G=S(n) show that A(n) (the subgroup of even
permuations) is also a normal subgroup of G.

Let N be a normal subgroup of G. Show that the order 2 element
in N is in the center of G if N and Z_4 are isomorphic.

A group G is a simple group if the only normal subgroups of G
are G itself and {e}. In other words, G is simple if G has no
non-trivial proper normal subgroups.
Algebraists have proven (using more advanced techniques than
ones we’ve discussed) that An is a simple group for n ≥ 5.
Using this fact, prove that for n ≥ 5, An has no subgroup of
order n!/4 .
(This generalizes HW5,#3 as well as our counterexample from...

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