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Normal Subgroup / Characteristic Subgroup Problem
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PostPosted: Fri May 20, 2005 10:13 pm    Post subject: Normal Subgroup / Characteristic Subgroup Problem Reply with quote
A problem from the Abstract Algebra, I.N.Herstein, 3/E:

"(one of) Middle-level Problems"

|G| = pm, p does not divide m, p : a prime.
H : a subgroup of G, |H| = p, H is normal to G.
==> H is characteristic to G

(In the book, more generalized case (|G| = p^n * m, |H| = p^n) exists.)

I'm self-studying, and only know about basic knowledge of group,
subgroup, Lagrange's theorem, homomorphism, and normal subgroup.
Clearly, Herstein may not expect no more complex theory. Any hint is
appreciated (since I worked over this problem several days, and tired
out). Thanks in advance. :)


Byunghyun Oh
Dept. of Mathematics, POSTECH
(but now in the military service of South Korea)
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Paul Sperry
science forum Guru


Joined: 08 May 2005
Posts: 371

PostPosted: Fri May 20, 2005 11:46 pm    Post subject: Re: Normal Subgroup / Characteristic Subgroup Problem Reply with quote
In article <1116634421.530434.230220@f14g2000cwb.googlegroups.com>,
<octaphial@gmail.com> wrote:

Quote:
A problem from the Abstract Algebra, I.N.Herstein, 3/E:

"(one of) Middle-level Problems"

|G| = pm, p does not divide m, p : a prime.
H : a subgroup of G, |H| = p, H is normal to G.
==> H is characteristic to G

(In the book, more generalized case (|G| = p^n * m, |H| = p^n) exists.)

I'm self-studying, and only know about basic knowledge of group,
subgroup, Lagrange's theorem, homomorphism, and normal subgroup.
Clearly, Herstein may not expect no more complex theory. Any hint is
appreciated (since I worked over this problem several days, and tired
out). Thanks in advance. Smile

Can you use Sylow's Theorem? (That is, do you know that H is the unique
subgroup of order p?)

--
Paul Sperry
Columbia, SC (USA)
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Byunghyun Oh
science forum beginner


Joined: 21 May 2005
Posts: 4

PostPosted: Sat May 21, 2005 3:46 am    Post subject: Re: Normal Subgroup / Characteristic Subgroup Problem Reply with quote
Answer : No. Sylow's theorem is far ahead from here, and this problem
is placed in the section of homomorpisms and normal subgroups. I have
imagined that "fact" (before your confirmation, "conjecture"), however,
from many examples I constructed and calculated. I am also convinced
that the proof will be trivial with it. Very thanks for your help
anyway!


Byunghyun Oh
Dept. of Mathematics, POSTECH
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Jim Heckman
science forum Guru Wannabe


Joined: 28 Apr 2005
Posts: 121

PostPosted: Sat May 21, 2005 7:38 am    Post subject: Re: Normal Subgroup / Characteristic Subgroup Problem Reply with quote
On 20-May-2005, octaphial@gmail.com
wrote in message <1116634421.530434.230220@f14g2000cwb.googlegroups.com>:

Quote:
A problem from the Abstract Algebra, I.N.Herstein, 3/E:

"(one of) Middle-level Problems"

|G| = pm, p does not divide m, p : a prime.
H : a subgroup of G, |H| = p, H is normal to G.
==> H is characteristic to G

(In the book, more generalized case (|G| = p^n * m, |H| = p^n) exists.)

I'm self-studying, and only know about basic knowledge of group,
subgroup, Lagrange's theorem, homomorphism, and normal subgroup.
Clearly, Herstein may not expect no more complex theory. Any hint is
appreciated (since I worked over this problem several days, and tired
out). Thanks in advance. Smile

OK, let's do it without Sylow's Theorems, per your other post. By
Langrange's Theorem, H is a maximal p-subgroup of G. Since H is normal,
any element g of G whose order is a power of p must be contained in H,
otherwise <H,g> would be a p-subgroup larger than H. So H contains all
elements of G whose order is a power of p, so ...

--
Jim Heckman
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Paul Sperry
science forum Guru


Joined: 08 May 2005
Posts: 371

PostPosted: Sat May 21, 2005 12:55 pm    Post subject: Re: Normal Subgroup / Characteristic Subgroup Problem Reply with quote
In article <1116634421.530434.230220@f14g2000cwb.googlegroups.com>,
<octaphial@gmail.com> wrote:

Quote:
A problem from the Abstract Algebra, I.N.Herstein, 3/E:

"(one of) Middle-level Problems"

|G| = pm, p does not divide m, p : a prime.
H : a subgroup of G, |H| = p, H is normal to G.
==> H is characteristic to G

(In the book, more generalized case (|G| = p^n * m, |H| = p^n) exists.)

[...]

OK. Since you say elsewhere that you don't have Sylow's Theorems let's
do it this way. As you'll see, we might as well do the generalization.

Let f be an automorphism of G, let K = f[H]. Our job is to show K < H.

Note the group G/H has order m. Also note that, for k in K, either
kH = H or kH has order a power of p. But, ...., so kH = H for all k in
K and thus K < H.

--
Paul Sperry
Columbia, SC (USA)
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Byunghyun Oh
science forum beginner


Joined: 21 May 2005
Posts: 4

PostPosted: Sat May 21, 2005 11:37 pm    Post subject: Re: Normal Subgroup / Characteristic Subgroup Problem Reply with quote
Thanks so much for your helps! I'll try it following your framework. :)


Byunghyun Oh
Dept. of Mathematics, POSTECH
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Ken Pledger
science forum Guru Wannabe


Joined: 04 May 2005
Posts: 268

PostPosted: Mon May 23, 2005 12:43 am    Post subject: Re: Normal Subgroup / Characteristic Subgroup Problem Reply with quote
In article <1116634421.530434.230220@f14g2000cwb.googlegroups.com>,
octaphial@gmail.com wrote:

Quote:
A problem from the Abstract Algebra, I.N.Herstein, 3/E:

"(one of) Middle-level Problems"

|G| = pm, p does not divide m, p : a prime.
H : a subgroup of G, |H| = p, H is normal to G.
==> H is characteristic to G

(In the book, more generalized case (|G| = p^n * m, |H| = p^n) exists.)

I'm self-studying, and only know about basic knowledge of group,
subgroup, Lagrange's theorem, homomorphism, and normal subgroup.
Clearly, Herstein may not expect no more complex theory. Any hint is
appreciated (since I worked over this problem several days, and tired
out)....

Think about orders of elements. The order of any element of H is
a factor of p. Apply to it any automorphism of G followed by the
natural homomorphism to G/H. You'll get a coset whose order in G/H is
still a factor of p, but by Lagrange's Theorem also a factor of m.
What follows? And how can you generalize it?

Ken Pledger.
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Google
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