subgroups and quotient groups of solvable groups are solvable _ project crazy project
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Subgroups and quotient groups of solvable groups are solvable
Prove that subgroups and quotient groups of solvable groups are solvable.
We prove some lemmas first.
Lemma 1: Let be a group and let with normal in . Then is normal in . Proof: Let . Then
because and .
Lemma 2: Let , , and be groups with normal in . Then is normal
in and there is an injective homomorphism .
Proof: Let be the natural projection. Certainly ,
and if , we have , and so . By the First Isomorphism
Theorem, the induced homomorphism is injective.
Lemma 3: Let be a group and such that is normal in and
is normal, and . Then is normal. Proof:
Let . Then .
Let be a solvable group. Then there exists a subnormal series
such that is abelian for all
.
Let be a subgroup. By Lemma 1 we have
. Moreover, by
Lemma 2 we have abelian. Thus issolvable.
Let be a normal subgroup. Now is normal byLemma 3, so that
. Moreover,
is abelian by the Third Isomorphism
Theorem. Hence is solvable.
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On these pages you will find aslowly growing (and poorlyorganized) list of proofs andexamples in abstract algebra.
No doubt these pages are riddledwith typos and errors in logic,and in many cases alternatestrategies abound. When youfind an error, or if anything isunclear, let me know and I will fixit.
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neal On July 11, 2011 at 11:09 pm Permalink | Reply
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« Finite abelian groups have subgroups of every order dividing the order ofthe group
Every finite group has a composition series »
By nbloomf, on April 11, 2010 at 11:00 am, under AA:DF, Incomplete. Tags: quotient group, solvablegroup, subgroup. 13 Comments
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why is N(K/N) = N again, thanks
I think you’re talking about Lemma 3. is not true, and I’m notseeing it here. Where is that step?
By the way, with the benefit of hindsight I can see that Lemma 3 is just part ofthe Lattice Isomorphism Theorem.
in lemma 3 where it says a(N(K/N)) = a(N/K), I’m just beginning to understand theidea of a coset
Eh… looking more closely, this solution is messed up. I’m going to have torewrite it later when I have more time. For now I’ll mark it ‘incomplete’.
So sorry for the inconvenience!
hey so is that operation well defined the one where aN K/N = aK/N,
i know aN is a coset of N and K/N is the set of all cosets ( can’t assume it has agroup structure since N is not assumed to be normal), so how do you multiplythem. thanks
I haven’t gotten around to fixing this one yet. It the (implicit, now explicit)assumption that , then is in fact normal in , and , so it works.
The ‘mod ‘ notation is a little misleading here. really means a set ofequivalence classes in with respect to the relation if and only if
.
Again, this will eventually be rewritten.
the indices are wrong, H_i is a subgroup of H_{i+1}
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nbloomf On November 28, 2011 at 10:30 am Permalink | Reply
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nbloomf On November 29, 2011 at 8:56 am Permalink | Reply
Gobi Ree On November 29, 2011 at 9:50 am Permalink | Reply
Thanks!
In lemma2, you can define and obviously .Then by the 1st isom thm, the result follows. We don’t have to check welldefinedness and injectivity.
Thanks that’s much better.
In the text, the definition of a solvable group requires just a subnormal series, nota composition series. And this definition is in wikipedia, too. Is it a mistake thephrase “composition series” in your post?
That was a mistake on my part. However, I think either works in this case, sinceany composition series of a solvable group will serve as a witness to solvability(i.e., have abelian intermediate quotients) and any witness to solvability (i.e.subnormal series with abelian intermediate quotients) can be refined to acomposition series.
I wrote an answer to the case of quotients groups. But it’s a little long and actuallyI wanted to write a post in wordpress, so I wrote it in my wordpress. (First post lol!)I’ll be very glad if you read it, thanks!http://gobiree.wordpress.com/2011/11/29/quotientgroupsofasolvablegrouparesolvable/
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