541 Day Lemma: If N is a normal subgroup of G and N is any subgroup of G then H N = HN = NH. Further if H is normal, NH is normal as well.

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Jocelin Crawford
6 years ago
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1 541 Day Section 34: Isomorphism Theorems: 1. First Isomorphism Theorem. Done! This is just another name for The Fundamental Homomorphism Theorem. 2. Definition: Let H and N be subroups of G. The join of H and N, denoted H N is the smallest subgroup of G containing HN. [Equivalently it is the intersection of all subgroups containing HN.] 3. Lemma: If N is a normal subgroup of G and N is any subgroup of G then H N = HN = NH. Further if H is normal, NH is normal as well. Proof. Suppose N is a normal subgroup of G and N is any subgroup of G. Let g be an element of NH. Then g = nh for some h in H and n in N. Since N is normal, g = h n for some h in H. So g is in HN. Thus NH HN. Now suppose that g is an element of HN. Then g = hn for some h in H. Note that g - 1 = n - 1 h - 1. That means that g - 1 is an element of NH. By the above we know that it is also an element of HN. Say, g - 1 = h1n1. But then g =( g - 1 ) - 1 = n1-1 h1-1. So that g is in NH. [That was a triple reversal proof! And it showed that although our definition of normality only says things like nh can be written as hn, normality also guarantees that things like hn can be written as n h. ] If we can show that HN is a subgroup then we are done. It is clearly nonempty since H and N are. If h1 n1and h2 n2 are in HN then n2h2 n1h1= n2 n3 h2h1 for some n3 in N. So the product n2h2 n1h1 is in HN. Now suppose that hn is in HN. It s inverse is n - 1 h - 1 which is in NH and hence in HN. So HN is a subgroup and hence it is the smallest subgroup containing HN. Therefore H N = HN = NH. Finally note that if g is in G and nh is in NH, then nhg = hgn = hgn = gh n = gn h. So NH is normal.

2 4. Theorem (Second Isomorphism Theorem): Let H be a subgroup of G and N be a normal subgroup of G. Then HN/N is isomorphic to H/(H N). Proof: First observe that any subgroup of a normal subgroup is also normal so H N is normal in G. Second, observe that a normal subgroup of a group is also normal in any subgroup containing it. So N is normal in HN and H N is normal in H. So both HN/N and H/(H N) are groups. Version 1 (Using First Isomorphism Theorem): Define ϕ: HN H/(H N) by ϕ(hn) = h(h N). Homomorphism: Let hn, h n be in HN. Then ϕ(hn h n ) = ϕ(hh n n ) for some n in N (by normality of N). So we have ϕ(hn h n ) = ϕ(hh n n ) = hh (H N) = h(h N) h (H N) = ϕ(hn)ϕ(h n). If hn is in the Kernel then h(h N) = e(h N), which implies that h is in H N. This means that h is in N, so then hn is in N. So the Kernel is contained in N. And if a product of the form hn is in N, then we know h must be in N which implies that ϕ(hn) = h(h N) = e(h N). So hn is in the Kernel. So the Kernel is N. Now the result follows from the first isomorphism theorem. Version2 (Direct): Define ϕ: HN/N H/(H N) by ϕ[(hn)n] = h(h N). Well defined? Suppose (hn)n = (h n )N. Note that this means that hn = h N. Which means that h - 1 h is an element of N. Which means that h - 1 h is an element of H N. Thus h(h N) = h (H N), so ϕ [(hn)n] =ϕ [(h n )N]. Thus this map is well defined. It is clearly everywhere defined. Onto: For any h(h N) in H/(H N), ϕ(he)n = h(h N). 1-1: Suppose ϕ [(hn)n] =ϕ [(h n )N]. Then h(h N) = h (H N), which means that h - 1 h is an element of H N. Then n - 1 h - 1 h n is in N. But n - 1 h - 1 h n = (hn) - 1 (h n ). So (hn)n = (h n )N. Homomorphism: ϕ[(hn)n(h n )N] = ϕ[(hn)(h n )N] = ϕ[(hh n n )N] = hh (H N) = h(h N) h (H N) = ϕ[(hn)n] ϕ[(h n )N].

3 5. Example: Let G = D8. Let H = {I, F} and N = {I, R 2 } Then HN = {I, R 2, F, FR 2 } and H N = {e} IN = {I, R 2 } FN = {F, FR 2 } R 2 N = {R 2, I} FR 2 N = {FR 2, F} So, HN/N = {IN, FN} I(H N) = {I} and F(H N) = {F} So, H/(H N) = {I(H N), F(H N)}. And now HN/N is obviously isomorphic to H/(H N)! Example Example: Let G = Z, H = az, and N = bz. HN = lcm(a, b)z H N =gcd(a, b)z HN/N = lcm(a, b)z/ bz H/(H N) = az/ gcd(a, b)z lcm(a, b)z/ bz = az/ gcd(a, b)z lcm(a, b)z/ bz = az/ gcd(a, b)z so lcm(a, b)z / bz = az / gcd(a, b)z Finally we get lcm(a, b)/b = a/gcd(a, b) Or lcm(a, b) = ab/gcd(a, b)

4 6. Third Isomorphism Theorem: Let H and K be normal subgroups of a group G with K H. Then G/H is isomorphic to (G/K)/(H/K). Proof: First observe that our definition of coset multiplication means that if H is normal in G then H/K is normal in G/K. Verification: For any gk in G/K and hk in H/K, gkhk = ghk = h gk for some h in H (by normality of H in G). And h gk = h KgK as desired. Version 1 (Using First Isomorphism Theorem) Proof 2 (Using First Isomorphism Theorem): Consider the map ϕ: G (G/K)/(H/K) by ϕ(g) = gk(h/k). This is a homomorphism since ϕ(ab) = (ab)k(h/k) = (akbk)(h/k) = ak(h/k) bk(h/k) = ϕ(a) ϕ(b). What is the kernel of this homomorphism? Suppose ϕ(g) = gk(h/k) = ek(h/k). Then (ek) - 1 gk is in H/K. But (ek) - 1 gk = (e - 1 K) gk = gk, and gk in H/K means that g is in H. So the kernel is contained in H. Further if h is in H, then ϕ(h) = hk(h/k). And since (ek) - 1 hk = hk is in H/K, then hk(h/k) = ek(h/k), so h is in the kernel. Since the kernel is H, we know by the First Isomorphism Theorem that G/H is isomorphic to (G/K)/(H/K). Version 2 (Direct) Define ϕ: G/H (G/K)/(H/K) by ϕ(gh) = gk(h/k). Well defined? Suppose ah = bh. Then a - 1 b is in H. Want to show that (ak) - 1 bk is in H/K. But (ak) - 1 bk = (a - 1 K)bK = (a - 1 b)k which is in H/K since a - 1 b is in H. So the map is well defined. It is clearly everywhere defined. Onto: Let gk(h/k) be an element in (G/K)/(H/K), then ϕ(gh) = gk(h/k). So the map is clearly onto. Homomorphism: ϕ(ah) ϕ(bh) = ak(h/k) bk(h/k) = akbk(h/k) = (ab)k(h/k) = ϕ(abh).

5 7. Example: Let G = D8. Let H = {I, R 2, F, FR 2 } and K = {I, R 2 } K is normal in D8 (it is the center) and so is H (it has index 2). G/H = {H, RH}, G/K = {K, FK, RK, FRK} H/K = {K, FK}, (G/K)/(H/K) = {H/K, R(H/K)} where R(H/K) = {RK, RFK} = {RK, FRK}

Math 4/541 Day 25. Some observations:

Math 4/541 Day 25. Some observations: Math 4/541 Day 25 1. Previously we showed that given a homomorphism, ϕ, the set of (left) cosets, G/K of the kernel formed a group under the operation akbk = abk. Some observations: We could have just