The most important isomorphism theorem, sometimes referred as the "fundamental homomorphism theorem" we will refer to as the "first isomorphism theorem".

**Theorem:**If G is a group with homorphism f into some other group then $$ G/\ker f \simeq f(G) $$. That is to say, G mod out the kernel of the morphism will give an isomorphism between the image of f.

**Proof:**First we need to know that the kernel is indeed a normal subgroup of G otherwise the factor group does not make sense. But that is very easy to show, if $$ a\in \ker f$$ and $$g\in G$$ then $$f(gag^{-1}) = f(g)f(a)f(g)^{-1} = f(g)\cdot 1\cdot f(g)^{-1} = 1$$. Now we make the observation that we can define a natural map $$\hat f$$ from $$G/\ker f$$ into $$f(G)$$ by the rule $$g(\bmod \ker f) \mapsto f(g)$$. This map is clearly well-defined because if h is chosen as another representative for a coset modulo the kernel then h = gk where k is some element in the kernel. And so $$f(h) = f(gk)=f(g)f(k)=f(g)$$. It is clear by definition that $$\hat f$$ is indeed as homomorphism. It is also not hard to see that it is one-to-one because the elements which are map to zero are precisely those in the kernel by definition. By construction of $$\hat f$$ we see that it is also onto. Thus, this natural morphism induces an isomorphism between the two groups. Q.E.D.

This is really an important result, really simple but very important, it essentially says that dividing out the kernel turns a homomorphism into a monomorphism (injective homomorphism).

Before we state the second isomorphism theorem we need some additional notation. If H and K are subgroups of a group G we will define HK (sometimes we will write H+K if we are working in an abelian group) to be the set of all products {hk| h in H and K in K}. In general, this set is not a group. But in the case when K is normal in G it is a group. Notice, that all the conditions of a subgroup are easy to verify for HK except closure. That is the one we will concentrate on. We need to show that $$(h_1k_1)(h_2k_2) = h_3k_3$$ to satisfy the closure condition. Just notice that $$h_2^{-1}k_1h_2=k_3$$ for some $$k_3$$ because K is normal. Thus, $$k_1h_2=h_2k_3$$. So, $$(h_1k_1)(h_2k_2) = (h_1h_2)(k_3k_2)$$ and the proof is complete because $$h_1h_2\in H$$ and $$k_3k_2 \in K$$. Now notice that the proof is similar if instead we had to proof KH was a subgroup. Let L be the smallest subgroup containing H and K. Then L must contain HK and KH. In the case if K is normal we have the HK and KH are subgroups. Therefore, L = HK and L = KH so HK = KH. This simple fact tells us that we need not concern ourselves how we write this product in the case when one of them is normal. Now we can state the second isomorphism theorem.

**Theorem:**Let G be group with subgroup H and normal subgroup K. Then $$HK/K\simeq H/(H\cap K)$$.

**Proof:**The proof of this is based on understanding the first isomorphism theorem correctly, it follows from the first. Before we begin notice that the two factor groups we be working with make sense because K is normal in HK and $$H\cap K$$ is normal in H. Define the function $$f: H\to HK/K$$ by the rule $$f(h) = hK$$. The map is clearly a homomorphism. It is also easy to see it is onto if you have the correct feel for factor groups. It is onto because anything in HK/K can be represented as a coset with element hk, but since we are taking modulo K the k element as if disappears and we are left with h. So everything in HK/K can be represented as a coset with element in H alone. However the map f is need not be one-to-one and so we do not have an isomorphism. However, if we mod out the kernel of f then we will have an isomorphism by the theorem above. The kernel of f is precisely the set $$H\cap K$$. The easiest way I think to see this is to consider the natural projection map $$\pi : HK \to HK/K$$. Notice that f is the restriction of the natural projection map to H. Since $$\ker \pi = K$$ it follows that $$\ker f = \ker \pi |_H = H\cap \ker \pi = H\cap K$$. Now by the first isomorphism theorem everything follows through nicely and the proof is complete. Q.E.D.

Before we state the third isomorphism theorem we need some remarks about factor groups. If H and K are both normal in G we can form the factor groups G/H and K/H under the hypothesis that H is sitting in K. Then K/H is a subgroup of G/H. It is also easy to see that K/H is in fact a normal subgroup of G/H. So we can form the factor group (G/H)/(K/H). This can be quite confusing in computational problems. A factor group is a set of sets. A factor group of a factor group will be a set of set of sets. But the next theorem says that this double factor group is actually quite simple up to isomorphism.

**Theorem:**Let H and K be normal subgroups of G with H sitting in K. Then we have $$(G/H)/(K/H)\simeq G/K$$.

**Proof:**Define the function $$f:G/H\to G/K$$ by $$f(g) =gK$$. Clearly this is well-defined since H sits in K and it is onto. The kernel are those $$g$$ such that $$gK=K$$. Thus, $$\ker f = K/H$$. Now apply the first isomorphism theorem to complete this proof. Q.E.D.

Now there happens to be a fourth isomorphism theorem. This is one is very rare, it hardly every comes up. The first theorem comes up all the time, you would have occasions to use it even if you are not an algebraist. The second one comes up from time to time, the third one is a little more rare but it still will come up. The fourth isomorphism theorem comes up in rare conditions. I have seen it applied only once.

Before we state the theorem we need some additional remarks about the product of groups HK. If both H and K are normal in G then it will follow that HK will remain normal in G. I will not go through all the details of this proof because it is very easy, just consider that $$g(hk)g^{-1} = (ghg^{-1})(gkg^{-1})$$.

**Theorem:**Let H and K be normal subgroups with H' and K' normal subgroups of H and K, respectively.

We have $$ H^|(H\cap K)/H^|(H\cap K^|) \simeq K^|(H\cap K)/K^|(H^|\cap K) $$

**Proof:**Of course implicity in the proof is that these factor groups make sense. But this will all be settled when we proof this result. Notice that $$H^|\cap K$$ and $$H\cap K^|$$ are normal subgroups of $$H\cap K$$. Therefore, if we define N to be the product of these two groups then N would be normal subgroup of $$H\cap K$$ by our remarks above. We will define a function $$f: H^|(H\cap K) \to (H\cap K)/N$$. By the rule that $$f(xy) = yN$$ where x in H' and y in $$H\cap K$$. We must show this map is well-defined. And then we must show it is a homomorphism. I will not do these proofs here. They are not hard, just follow the definitions. Be sure to remember to use the fact that H' is normal in H, it is essential in the proof. It is not hard to see that $$\ker f = H^|(H\cap K^|)$$. Therefore, we have proved $$H^|(H\cap K)/H^|(H\cap K^|) \simeq (H\cap K)/N$$. By a similar argument we can define a $$g: K^|(H\cap K) \to (H\cap K)/N$$ which will give us, $$K^|(H\cap K)/K^|(H^|\cap K)\simeq (H\cap K)/N$$. Thus, the two groups we want to prove are isomorphic are isomorphic to the same group, it follows they are isomorphic to eachother. Q.E.D.

Notice that in in the second to fourth isomorphism theorems we have constantly used the first. Master the first, it is the most important and useful one to have available in algebra.

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