# Why “normal subgroups occur as kernels of homomorphisms” is a big deal?

I already knew that normal subgroups where important because they allow for quotient space to have a group structure. But I was told that normal subgroups are also important in particular because they are the only subgroups that can occur as kernels of goup homomorphisms. Why is this property a big deal in algebra?

• Well, because kernels are a big deal, I guess ;) – NeitherNor Jul 10 at 13:04
• Normal subgroups are also a "big deal", because groups without non-trivial normal subgroups are so important. Being kernel of a group homomorphism is just equivalent to being normal, so nothing new. – Dietrich Burde Jul 10 at 13:08
• Practically the same issue. Quotient spaces correspond one to one with the kernels of the natural maps that are involved. – drhab Jul 10 at 13:14
• Associated with a group $G$ we often get (or can cook up) its action on some set $X$. This means that we have a non-trivial homomorphism from $G$ to $Sym(X)$. When cardinality considerations dictate that the said homomorphism cannot be injective, we can conclude that $G$ has a non-trivial normal subgroup. This theme recurs when proving that a group of certain order cannot be simple. – Jyrki Lahtonen Jul 10 at 13:17
• Because essentially there is a severe poverty of big deals in Algebra, so you get what can. – uniquesolution Jul 10 at 13:17

I suppose this is more of opinion-based question. If I interpret your question as "Why is it a big deal that the subgroup determined by the kernel of a homomorphism must be normal?", then the answer is "It's not really a big deal. That's trivial." Instead, if I interpret your question as "Why is it a big a deal that normal subgroups arise as kernels of homomorphisms?" then that's another story.

There are many situations where you can prove something interesting about (normal) subgroups by being clever with kernels of homomorphisms. Here are two simple examples.

Fact 1: If $$G$$ is a group and $$H$$ is a subgroup of index $$n$$. Then $$H$$ contains a normal subgroup $$K$$ of $$G$$ with index at most $$n!$$.

Proof: Let $$X$$ be the set of left cosets of $$H$$. Then any element of $$g$$ determines a permutation of $$X$$ in which each coset is multiplied on the left by $$g$$. Define the function $$f:G\to S_{X}$$ sending $$g$$ to its associated permutation. (Here $$S_{X}$$ is the group of permutations of $$X$$.) It is easily checked that $$f$$ is a homomorphism. Let $$K$$ be the kernel. Then $$K$$ is a normal subgroup of index at most $$|S_{X}|=n!$$, and it is easy to check that $$K$$ is contained in $$H$$.

In the previous proof, the definition of $$f$$ implies that $$K$$ is the intersection of all conjugates of $$H$$. So $$K$$ is the "largest" normal subgroup of $$G$$ that is contained in $$H$$.

Comment (added later). Continuing from the last sentence, set $$K=\bigcap_{g\in G}gHg^{-1}$$. So $$K$$ is the largest normal subgroup of $$G$$ that is contained in $$H$$. And the proof above shows that it has index at most $$n!$$. But one my try the following more direct approach. Choose left coset representatives $$g_{1},\ldots,g_{n}$$ for $$H$$. It is not hard to see that $$K=\bigcap_{t=1}^{n}g_{t} Hg_{t}^{-1}$$. Also, any conjugate of $$H$$ still has index $$n$$. So $$K$$ is an intersection of $$n$$ subgroups of $$G$$ each of index $$n$$. Now the general formula $$[G:H_1\cap H_2]=[G:H_1]\cdot [G:H_2]$$ tells us that $$K$$ has index at most $$n^n$$. So this proves the Fact without using homomorphisms but with a worse bound on the index.

Fact 2: Suppose $$G$$ has order $$2n$$ where $$n$$ is odd. Then $$G$$ has a normal subgroup of size $$n$$.

Proof: Any element of $$g$$ determines a permutation of $$G$$ via multiplication on the left. So we get a map $$\varphi:G\to S_{G}$$ which is a homomorphism. Let $$\psi:S_G\to C_2$$ be the map that sends a permutation in $$S_G$$ to $$0$$ if and only if it is even. (Here $$C_2$$ is the cyclic group of size $$2$$.) Compose these to get a homomorphism $$f: G\to C_2$$. By Cauchy's Theorem, $$G$$ has an element $$x$$ of order $$2$$. So $$\varphi(x)$$ is a permutation of order two with no fixed points. So $$\varphi(x)$$ is a product of $$n$$ disjoint transpositions. Since $$n$$ is odd, we have $$\psi(\varphi(x))=1$$. So $$f$$ is surjective. If $$K$$ is the kernel of $$f$$ then $$G/K$$ is isomorphic to $$C_2$$. So $$K$$ has index $$2$$, i.e. size $$n$$.

The last fact tells us there are no simple groups of size $$2n$$ if $$n$$ is odd and bigger than $$1$$.

By the way, the last proof used Cauchy's Theorem for $$p=2$$. I know it's got nothing to do with your main question, but here's a cute proof. Suppose $$G$$ is a group with even order. We aim to find an element of order $$2$$. Let $$X$$ be the set of elements of order greater than $$2$$. Then no element in $$X$$ is equal to it's inverse, so we can partition $$X$$ into sets of size $$2$$ by putting each element with its inverse. So $$|X|$$ is even which means $$|G\setminus X|$$ is even. Since $$G\setminus X$$ contains the identity, it must contain at least one more element, which has order 2.

Let $$G$$ be a group. Since the normal subgroups of $$G$$ coincide, as a set, with the subgroups that appear as kernels of homomorphisms with domain $$G$$, the normal subgroups are exactly the subgroups of $$G$$ that appear as the left object in short exact sequences of the form $$1\longrightarrow N\longrightarrow G\longrightarrow K\longrightarrow 1.\qquad(*)$$ The point is that once you have an exact sequence $$(*)$$ the group $$G$$ can be reconstructed out of $$N$$ and $$K$$ plus some extra-combinatorial data (technically, cohomological data depending only on $$K$$ and $$N$$).

Now suppose that $$G$$ is finite. Then, the isomorphism theorem tells you that $$|G|=|N|\cdot|K|$$, i.e. the group $$G$$ can be reconstructed out of smaller groups plus some additional data depending only on those smaller groups.

If you have a list of finite groups that contain no normal subgroups (these groups are called simple) the above sets the very first step towards the goal of reconstructing all finite groups.

When $$G$$ is not finite the property has still some interest, for instance when studying representations of $$G$$, i.e. homomorphisms of the kind $$G\longrightarrow{\rm GL}(V)$$ where $$V$$ is some vector space.