“If you are always trying to be normal, you will never know how amazing you can be.”
―Maya Angelou
Sometimes it is convenient to ignore a certain subobject of a given object and to focus on the remaining properties. Formally, this is done by taking quotients. In contrast to the theory of vector spaces, where the quotient of any vector space by any subspace again naturally forms a vector space, we have to be a little bit more careful in the world of groups. Only special subgroups lead to quotient groups:
Definition3.17.Normal Subgroup.
A subgroup \(N\) of a group \(G\) is normal in \(G\text{,}\) written \(N\nsg G\text{,}\) if \(gNg^{-1} = N\) for all \(g \in G\text{.}\)
Discussion3.1.Anything but Normal.
Come up with a better name for normal subgroups than ’normal subgroup’. Give it some oomph. Make it mean something.
Hint.
Here are some options the author has recently become partial to:
Semiabelian: Yes, its more letters, but so are semigroup, semisimple, semidirect, and semicircle. Imagine if we called all those things normal too! It would be a nightmare. And we would deserve it for our hubris.
Ideal subgroup: Look, its shorter! Normal subgroups and ideals parallel ideas in group and ring theory, and relabeling normal groups in this way would cement these similarities.
However, there are many ways to characterize normal subgroups.
Theorem3.18.Equivalent Normal Characterizations.
Let \(N\) be a subgroup of a group \(G\text{.}\) The following are equivalent:
\(\displaystyle N \nsg G\)
\(gNg^{-1}\subseteq N\) for all \(g \in G\text{.}\)
\(gN = Ng\) for all \(g \in G\text{.}\)
The equivalence relations on \(G\) determined by the left and right cosets of \(H\) coincide.
Arguably, the most important and / or intuitive characterization of normal subgroups comes from their relation to homomorphisms. Indeed, though we currenly lack the tools to prove it, Theorem 3.40 tells us that a subgroup is normal if and only if it is the kernel of some homomorphism. Thus, in addition to its necessity in the definition of quotients, a normal subgroup is a simple and unique way to characterize any homomorphism.
Example3.19.Normal Subgroups.
The trivial subgroups \(\{e\}\text{,}\)\(G\) of a group \(G\) are normal subgroups of \(G\text{.}\)
Any subgroup of an abelian group is a normal subgroup.
The commutator subgroup \(G'\) from Example 2.11 is normal in \(G\text{.}\)
The set \(V=\{e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)\}\) is normal in \(S_4\text{.}\)
Theorem3.20.Closure Properties of Normal Subgroups.
Normalility is preserved under taking intersections and preimages.
Exercise3.21.Normality is not Transitive.
Show that being a normal subgroup is not a transitive relation. In other words, find groups \(K\leq H\leq G\) such that \(K\nsg H\) and \(H\nsg G\text{,}\) but \(K\not\nsg G\text{.}\)
Hint.
\(S_4\) is a good place to start routing around in.
From Example 3.19 we have \(V \nsg S_4\text{.}\) The group \(V\) has order \(4\text{,}\) making it ableian from Exercise 1.30. Thus every subgroup of \(V\) is normal in \(V\) by Example 3.19.
Now consider the subgroup \(H = \{e,(12)(34)\}\text{.}\) But \(H\) is not normal in \(S_4\text{,}\) since for example
If \(\varphi:G\to H\) is a group homomorphism and \(K\nsg H\) then the preimage of \(K\text{,}\)\(\varphi^{-1}(K)=\{g\in G | \varphi(g)\in K\}\text{,}\) is a normal subgroup of \(G\text{.}\)
Exercise3.23.Images not Normal in General.
Show that if \(\varphi:G\to H\) is a group homomorphism and \(L \nsg G\) then \(\varphi(L)\) need not be a normal subgroup of \(H\text{.}\)
Here is a theorem that will prove exceedingly useful all the way down in Chapter 6.
Theorem3.24.Unique Order Subgroup is Normal.
Let \(H\leq G\) be the only subgroup of order \(n\text{.}\) Then \(H\) is normal.
We also have some criteria for when a generated subgroup is normal.
Exercise3.25.Normality and Generated Subgroups.
Let \(G\) be a group and \(S\) a subset of \(G\text{.}\) Suppose \(H=\langle S \rangle\text{.}\) Prove \(H\nsg G\) if and only if \(gsg^{-1}\in H\) for every \(s\in S\) and \(g\in G\text{.}\)
Or, if we’re tired of generating subgroups that aren’t as normal as we’d like them to be, we can generate a normal subgroup explicitely.
Exercise3.26.Generated Normal Subgroup.
Let \(G\) be a group and let \(A \subseteq G\) be a set. The normal subgroup generated by \(A\), denoted \(\langle A \rangle ^N\text{,}\) is the intersection of all the normal subgroups of \(G\) that contain \(A\text{.}\)
Show that \(\langle A \rangle ^N\) is a normal subgroup of \(G\text{.}\)
Show that the elements of \(\langle A \rangle ^N\) can be described as
\begin{equation*}
\langle A \rangle ^N= \{ g_1a_1^{i_1}g_1^{-1} \dots g_ma_m^{i_m}g_m^{-1} \mid m \geq 0, a_j \in A, g_j \in G, i_j \in \{1,-1\}, \forall 1\leq j\leq m\}.
\end{equation*}
We conclude this section by taking a sneak peak at the interactions between the index of a subgroup and its normality.
Proposition3.27.Subgroup of Index \(2\) is Normal.
Show that any subgroup of index two is normal. This means: show that if \(G\) is a group, \(H\) is a subgroup and \([G:H]=2\text{,}\) i.e the number of left (or right) cosets of \(H\) in \(G\) is two, then \(H\) is normal.
Proof.
Coming soon to an OER near you!
Remark3.28.
This is a specific case of Theorem 5.16, a much more powerful theorem that we will prove later on, but we can tackle this smaller piece with the tools we have now.
Exercise3.29.Converse to Lagrange’s Theorem is False.
Prove the converse to Lagrange’s Theorem is false: find a group \(G\) and an integer \(d>0\) such that \(d\) divides the order of \(G\) but \(G\) does not have any subgroups of order \(d\text{.}\)
Hint.
Take \(G=A_4\text{.}\)
Solution.
Consider \(G=A_4\text{,}\) and note that \(|G|=12\text{.}\) Suppose by way of contradiction that \(H\) is subgroup of \(G\) such that \(|H|=6\text{.}\) Notice that \([G:H]=2\text{,}\) the smallest prime dividing the order of \(G\text{,}\) making \(H\) normal in \(G\text{.}\)
As there are eight \(3\)-cycles in \(G\text{,}\) there exists some \(3\)-cycle, \(\sigma\text{,}\) such that \(\s\not\in H\text{.}\) Consider then \(H, \s H,\) and \(\s^2 H\) in \(G/H\text{.}\) Since \(|G/H|=\frac{|G|}{|H|}=2\text{,}\) it must be the case that either \(\s^2H=\s H\) or \(\s^2H=H\text{.}\)
If \(\s^2 H=H\) then \(\s^2\in H\text{.}\) As \(|\s|=3\) we have \(\s^2=\s\inv\text{,}\) but as \(H\) is a subgroup this would mean \((\s\inv)\inv=\s\in H\text{,}\) which is not the case.
If \(\s^2 H=\s H\) then \(\s\inv\s^2\in H\text{,}\) but \(\s\inv\s^2=\s\text{,}\) and so we have a contradiction. Thus \(H\) cannot exist, and \(G\) has no subgroup of order \(6\text{.}\)
Qual Watch.
Providing such an example as in Exercise 3.29 was Part (b) of [cross-reference to target(s) "jan-2020-1" missing or not unique] on the [cross-reference to target(s) "jan-2020" missing or not unique] qualifying exam.
