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Course: MATH 375, Fall 2009
School: Hobart and William...
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375 Math Week 9 9.1 Normal Subgroups We have seen the two pieces of the folowing result earlier in the course. THEOREM 1 Let H be a subgroup of G and let g 2 G be a xed element. a The set gHg,1 is a subgroup of G. b jH j = jgHg,1j. Make sure that you understand this. Try proving this right now before reading the proof below. Begin by using the two-step test to prove the rst part and then show that the mapping : H...

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375 Math Week 9 9.1 Normal Subgroups We have seen the two pieces of the folowing result earlier in the course. THEOREM 1 Let H be a subgroup of G and let g 2 G be a xed element. a The set gHg,1 is a subgroup of G. b jH j = jgHg,1j. Make sure that you understand this. Try proving this right now before reading the proof below. Begin by using the two-step test to prove the rst part and then show that the mapping : H ! gHg ,1 is an isomorphism which proves the two sets have the same number of elements. PROOF A few weeks ago we saw that if : G1 ! G2 is a group homomorphism and H G1 , then H G2 . So for a xed g 2 G, de ne : G ! G by x = gxg ,1 . We will show that is an isomorphism and this will prove both parts of the theorem. is a homomorphism because a b = gag ,1gabg ,1 = g abg ,1 = ab: is injective because by cancellation, a = b gag ,1 = gbg ,1 a = b: is surjective because for any x 2 G codomain, g ,1 xg 2 G domain since G is a group closed. But g ,1 xg = g g ,1xg g ,1 = x: Then observe that part a of the theorem is true because H = gHg,1 and part b is true because is an isomorphism: jH j = j H j = jgHg,1j. We now single out certain subgroups H for special attention: 1 2 DEFINITION 2 Math 375 Let H be a subgroup of a group G. Then H is normal in G if gH = Hg for all g 2 G left and right cosets are equal. This is denoted by writing H G. At this point, it is not entirely clear why you would want to look at normal subgroups. But we will see that if H G, then the set of all left cosets of H is itself a group in a natural way. By next class, we should be able to prove this. For the moment, let's review what we done in lab and previous classes where we looked at left and right cosets. We saw that: An Sn , S2 is not normal in S3, and J Q8 . The next theorem gives us a criterion that makes it possible to show a H G without having to actually compute gHg ,1 for every g in G. THEOREM 3 Let H be a subgroup of G. Then the following three conditions are equivalent. i H is normal in G, that is, gH = Hg for all g 2 G; ii for all g 2 G, we have H = gHg ,1; iii ghg ,1 2 H for every g 2 G and every h 2 H , that is, gHg ,1 H . PROOF Notice that i ii was done as part of the coset property theorem. To do ii iii is easy: Since gHg ,1 = H , then we must have ,1 2 H for all g 2 G and all h 2 H . ghg Finally, for iii ii: This is really just the fact that the map in Theorem 1 is onto. ghg ,1 2 H for all g 2 G and h 2 H implies gHg,1 H . Now we need to show that for any g 2 G, H gHg,1. That is, given any g 2 G, we must show that any h 2 H can be written in the form h = gh0 g ,1 , for some h0 2 H . In particular, let x = g ,1 . Now set h0 = xhx,1 . Then by assumption iii, we know that h0 2 H . So h = gg,1hgg,1 = gxhx,1g = gh0g,1 2 gHg,1; which is what we wanted to show. EXAMPLE a Use condition iii to show that SLR; n GLR; n. b Along the same lines: Let GL+ R; be the the set of matrices with positive determinant. This is a subgroup of GLR; n. You might check that. Show that GL+ R; n GLR; n. THEOREM 4 Let H be a subgroup of G such that G : H = 2. Then H G. EXAMPLE 2 PROOF THEOREM 5 The two left cosets are H and G , H . But so are the right. Now give another proof that GL+R; n GLR; n. C G G. Further, if H is a subgroup of C G, then H G. 9.1 Normal Subgroups PROOF COROLLARY 6 3 Let H be a subgroup of C G. Then 8g 2 G and 8h 2 H , we have ghg,1 = hgg,1 = h 2 H . So by condition of iii the Theorem 3, H G. Let H be a subgroup of an abelian group G. Then H G. Note C G = G. Let H be a subgroup of a cyclic group G. Then H G: PROOF COROLLARY 7 PROOF G is cyclic so it is abelian. There are other simple criteria for normality which depend more on the subgroup H , than the group G. Here's one that is sometimes useful. If H is a subgroup of G and no other subgroup of G has the same order as H , then H is normal. We know that for any g 2 G that gHg ,1 is a subgroup of G and that jH j = jgHg,1j. But there are no subgroups of the order of H in G except H itself. So we must have gHg ,1 = H for all g . Let G = S3 Z5 . Then clearly G is not abelian since S3 is not. Let COROLLARY 8 PROOF EXAMPLE 3 H = 1; 1 = f1; 1; 1; 2; 1; 3; 1; 4; 1; 0g: Show that H G. SOLUTION Notice that G : H = 6 6= 2 and G is not abelian so none of the obvious results on normality apply. However if we can show that H is the only subgroup of G of order 5, then H G. We can do this by using a proof by contradiction. Suppose some other subgroup K also had order 5. Then K is cyclic why?, thus K = ; n why? where 2 S3 and n 2 Z5 . We know that 5 = j ; n j = lcmj j; jnj: What are the possible orders of ? What are the possible orders of n? Show that the only way the lcm to be 5 is for j j = 1 and jnj = 5. Thus = 1 and n can be 1, 2, 3, or 4 why?. In any of these cases K = 1; n = H . Thus H G. Can you also do this same problem above by showing H is a subgroup of C G? For another illustration of this last corollary, see p. 187 39. EXAMPLE 4 EXAMPLE 5 4 DEFINITION 9 Math 375 Let H be a subgroup of G. Let N H = fg 2 G j gHg,1 = H g: N H is called the normalizer of H in G. EXAMPLE a b LEMMA 10 Find the normalizer of another subgroup of a nonabelian group. The normalizer of a subgroup is important because: a N H is a subgroup of G; b H N H ; c H N H . Use the one-step test: Let a; b 2 N H . This means that aHa,1 = H and bHb,1 = H . The latter implies that H = b,1 Hb. We must show ab,1 2 N H , that is, ab,1H ab,1,1 = H . But ab,1 H ...

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X-Apparently-To: wishih@yahoo.com via web20201.mail.yahoo.com; 19 Nov 2001 09:25:36 -0800 (PST) X-RocketRCL: 2668;1;3620900138 X-Track: 1: 40 Date: Mon, 19 Nov 2001 11:27:49 -0600 From: Scott Henninger &lt;scotth@cse.unl.edu&gt; X-Mailer: Mozilla 4.7
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To: Scott Henninger &lt;scotth@cse.unl.edu&gt; From: William Shih &lt;wishih@yahoo.com&gt; Subject: USC MBASE cs577 Domain Corrupted Cc: Bcc: Scott,This is to follow up on Ben's earlier email about the problem with the MBASE domain. As a recap, there
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To: Scott Henninger &lt;scotth@cse.unl.edu&gt; From: William Shih &lt;wishih@yahoo.com&gt; Subject: Access Control Requirements Cc: Bcc: Scott,I apologize for the lateness of this email. I understand that you've probably been working on this for a whil
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To: Scott Henninger &lt;scotth@cse.unl.edu&gt; From: William Shih &lt;wishih@yahoo.com&gt; Subject: Cases not in tree in Rule Manager Cc: Bcc: Scott,I was working on adding more cases to Rule 0 when I noticed that some cases didnot show that there wer
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X-Apparently-To: wishih@yahoo.com via web20210.mail.yahoo.com; 22 Jan 2002 11:13:27 -0800 (PST) X-RocketRCL: 2454;1;3153585994 Date: Tue, 22 Jan 2002 11:13:26 -0800 (PST) From: Mohamed Keshk &lt;mkeshk1@yahoo.com&gt; Subject: USC Access Control Require
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To: Scott Henninger &lt;scotth@cse.unl.edu&gt;Subject: Add users to BORE Scott,I think I forgot to have you add two of our team members to the BORE system when we met together. Also, this semester, we've had one more person come onto our project.
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To: Ryan Kinworthy &lt;ryank@cse.unl.edu&gt;Subject: Re: USC Access Control RequirementsCc: Mohamed Keshk &lt;mkeshk1@yahoo.com&gt;, Scott Henninger &lt;scotth@cse.unl.edu&gt; Ryan,Thanks for your response. I understand about being busy, as we are alsovery b
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To: Scott Henninger &lt;scotth@cse.unl.edu&gt;Subject: Error Found in BORE - Domain Shows Only One CaseCc: CSCI577 Scott,This is a follow-up email from Ben Luc.We discovered a problem with BORE that reoccurred from last semester. When we were
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To: Scott Henninger &lt;scotth@cse.unl.edu&gt;Subject: BORE 4.x.x Client - Defect (Creating project adds cases to domain)Cc: Team20, Team21X-Attachments: E:\Documents\College Classes\CSCI 577b\newproj.gif; E:\Documents\College Classes\CSCI 577b\domaina
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To: Scott Henninger &lt;scotth@cse.unl.edu&gt;Subject: Keep MS Access VersionCc: Team20, Team21 Scott,As our Core Capability Demonstration milestone is coming up, and we're puttingin hours trying to get everything just right. I was wondering if we
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To: Scott Henninger &lt;scotth@cse.unl.edu&gt;Subject: BORE 4.10.48 - Defect (Related Cases overpopulated)Cc: Team20, Team21 Scott,I was reviewing some of my cases, and I noticed that there were a long list ofrelated cases with each case. Upon fu
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To: Scott Henninger &lt;scotth@cse.unl.edu&gt;Subject: BORE 4.10.18 - Defect (BORE not loading in Netscape)Cc: Team20, Team21 Scott,Katherine found this error when she was trying to use BORE with Netscape. She was not able to load even the login sc
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To: Scott Henninger &lt;scotth@cse.unl.edu&gt;Subject: Core Capability Demonstration ResultsBcc: William Shih &lt;wshih@mail.com&gt; Scott,As you know, we were approaching our Core Capability Demo milestone veryquickly, and many things were happening.
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To: Scott Henninger &lt;scotth@cse.unl.edu&gt;Subject: BORE 5.0.2 - Delete Case causes case to load Scott,When I click on a case in the Case Manager and click on the Delete Case button,the case loads instead, and the case opens to the Deviation Rat