# 另一條 Group theory

If H and K are subgroup of G with K normal in G, show that

(a) HK is a subgroup of G and (H∩K)&lt;| H

(b) φ(hk)=(H∩K)h is a homomorphism HK to H/(H∩K)

(c) φ is onto H/(H∩K)

(d) ker(φ) is K

(e) φ is not multiple-valued [ if h1k1=h2k2, show that (H∩K)h1=(H∩K)h2 ]

(f) HK/K ≈ H/(H∩K)

### 1 個解答

• 貓朋
Lv 5
1 十年前
最愛解答

I will still refer to the book, Algebra , by Michael Artin.

In fact, part(a) is just the results of p63(8.6)(b) and p60(7.1) , part(f) is just a direct application of First Isomorphism Theorem, p68(10.9).

(a) The trick is to show HK=KH first.

∀h ∈H and for ∀k ∈K, hk = (hkh-1)h ∈KH since hkh-1 ∈K as K is normal.

Similarly, kh= h(h-1kh) ∈HK and this shows that HK=KH

Now, h1k1h2k2 = h1h'2k'1k2 ∈HK (Since HK=KH, there exists h'2 and k'1 such that k1h2 = h'2k'1 )

Since e ∈H and e ∈K, e= ee ∈HK

Besides, (hk)-1 = k-1h-1 ∈KH =HK

This has shown that HK is a subgroup of G.

Since intersection of subgroup is still a subgroup and (H∩K) ⊂ H , (H∩K) < H. It remains to show that (H∩K) is normal to H.

For any g ∈(H∩K) and h ∈H, hgh-1 ∈K since K is normal.

Besides, hgh-1 ∈H since g ∈(H∩K) ⊂ H .

This means that hgh-1 ∈(H∩K) and hence (H∩K) is normal to H.

(b)

φ(h1k1h2k2)

= φ(h1h2h2-1k1h2k2)

= φ(h1h2k3k2) ...since K is normal, ∃k3 =h2-1k1h2

= (H∩K)h1h2

= (H∩K)h1 (H∩K)h2

= φ(h1k1) φ(h2k2)

(c) ∀s ∈H/(H∩K) , ∃h ∈H such that s= (H∩K)h

Now, h= he ∈HK

Hence, φ is onto.

(d) ∀k ∈K ,

φ(k)= φ(ek)= (H∩K)e ⇒ K ⊂Ker(φ)

On the other hand, ∀g ∈Ker(φ) ⊂HK, we have φ(g)= (H∩K)e and g=hk for some h ∈H and k ∈K . Now,

φ(hk)= φ(g)= (H∩K)e ⇒ h ∈(H∩K) ⊂K

Hence, g=hk ∈K

(e) If h1k1=h2k2 , then

(H∩K)h1=φ(h1k1) =φ(h2k2) =(H∩K)h2

(f) By First Isomorphism Theorem,

HK/Ker(φ) ≈ Im(φ)

i.e. HK/K ≈ H/(H∩K) by part (c), (d) and (e).

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