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As the define goes:

A subgroup $N$ of a group $G$ is called a normal subgroup if it is invariant under conjugation; that is, for each element $n$ in $N$ and each $g$ in $G$, the element $gng^{−1}$ is still in $N$.

My Question is: Can anyone give me an intuitive explaination or an example of this concept? Why it is very important in algebra?

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    $\begingroup$ This is the way I think of things: A normal subgroup of an arbitrary group is the "correct" generalization of an arbitrary subgroup of an abelian group, with respect to factoring (taking quotients). $\endgroup$ – Eric Stucky Nov 10 '14 at 7:36
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    $\begingroup$ This may be a duplicate of Intuition behind normal subgroups. Or perhaps not, since this question seems to be aimed at a lower level than that one. $\endgroup$ – Eric Stucky Nov 10 '14 at 7:37
  • $\begingroup$ If you view subgroups of a group acting on some space as (possible) stabilizers, normal subgroups are those which stabilize not only points, but whole orbits. In that sense, they are global versions of subgroups. $\endgroup$ – k.stm Nov 10 '14 at 19:15
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    $\begingroup$ Read this: gowers.wordpress.com/2011/11/20/… math.ucr.edu/home/baez/normal.html $\endgroup$ – ante.ceperic Nov 11 '14 at 8:18
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Let's take a look at the group of rotations of cube. It has a subgroup of rotations around vertical axis. This subgroup (let's call it $A1$) has 4 elements: rotate the cube for 0, 90, 180 or 270 degrees.

There is another subgroup: rotation around one of horizontal axes. Let's call it $A2$.

Subgroups $A1$ and $A2$ are obviously different. But still they look so very much alike! If there was someone else looking at our cube from different angle he could even fail to understand my descriptions of $A1$ and $A2$ "correctly" and confuse $A1$ with $A2$.

This is because $A1$ and $A2$ are conjugated. The $g x g^{−1}$ actually means "look at $x$ from another point of view", and $g$ defines this "point of view".

Subgroup is normal if it is very "symmetric". No matter from which point you look at the whole group $G$ the subgroup $N$ remains at place.

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    $\begingroup$ Nice but unsufficient : for a proper grasp of the idea, a second example and a NONE example would be nice. $\endgroup$ – Jérôme JEAN-CHARLES Aug 10 '15 at 12:46
  • $\begingroup$ Interesting. Does this extend to infinite-dimensional spaces, like let's say a change of numeraire in a time series? $\endgroup$ – isomorphismes Sep 20 '15 at 21:46
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Quotients of groups are only well defined if we take the quotient over a normal subgroup. Another way of writing your definition is that $N$ is normal iff $gN = Ng$, so the set of left cosets equals the set of right cosets, which makes the quotient group $G/N$ well-defined. That is, in my opinion the key reason to be interested in them. Otherwise, normal subgroups keep popping up literally everywhere in group theory. For example, when we look at solvable groups (Galois theory), they are important.

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  • $\begingroup$ Every subgroup's cosets partition the group. So the quotient group, or the set of cosets, can be defined as a group in many different ways. However, the composition rule for cosets is well defined with respect to representative elements iff the subgroup is normal, which, to clarify, is what I think Johanna means by G/N being well defined. $\endgroup$ – David Warren Katz Jun 27 '16 at 2:57
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A congruence in a group $G$ is an equivalence relation $\equiv$ in $G$ that is compatible with the operation of $G$: $$ a \equiv b, \ a' \equiv b' \implies aa '\equiv bb' $$ The quotient $\overline G = G\,/\equiv$ is then a group.

It is easy to prove that, when $\equiv$ is a congruence in $G$, the equivalence class of $1$ is a normal subgroup $N$ of $G$ and the equivalence classes are the cosets of $N$.

Conversely, if $N$ is a normal subgroup of $G$, then the relation defined by $a \equiv b$ if $a^{-1}b \in N$ is a congruence relation in $G$ whose equivalence classes are the cosets of $N$.

So, normal subgroups are natural objects when you consider congruences. The other equivalent characterizations of normality, such as $aN=Na$ and invariance under conjugation, follow easily.

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