Re: algebra with cyclic and normal.
- From: Kira Yamato <kirakun@xxxxxxxxxxxxx>
- Date: Wed, 27 Dec 2006 04:33:12 -0500
On 2006-12-27 04:09:20 -0500, "mina_world" <mina_world@xxxxxxxxxxx> said:
Hello sir~
Teachers,
In the fisrt, I thank you for teaching me mathematics for 2006 year.
I hope you to teach me mathematics in next year too.
I'm no teacher. Just a fellow student of mathematics helping each other.
Anyway, my problem is...
G : group.
H : normal cyclic subgroup of G.
Show that all subgroups of H are normal of G.
H = <h> is normal in G means
for all g in G, for all integer n, there exists an integer m
such that
(1) g^{-1} h^n g = h^m.
Suppose K is a subgroup of H. Then
K = <h^k>
for some integer k. So, raising (1) on both side to the power of k gives
g^{-1} h^{nk} g = h^{mk}.
So,
g^{-1} K g is contained in K.
So, K is normal in G.
--------------------------------------------------
is this possible ?
In fact, I know that
H = <p_1> = {p_0, p_1, p_2, p_3} normal cyclic subgroup of group D_4.
and {p_0} and {p_0, p_2} are really normal of D_4.
Let N be a subgroup of H.
so, N is cyclic subgroup of H. (H : cyclic)
Since H is abelian, N is normal subgroup of H.
Since H is cyclic, H = <a> , (a in G).
so, N = <a^j> form. (j in N).
I must show that g.a^j.(g^-1) in N.
But, I don't know well.
so, I need your advice.
--
-kira
.
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