= Solution
A <deck transformation> of a covering $p:\widetilde X\to X$ is a <homeomorphism> $d:\widetilde X\to\widetilde X$ with $p\circ d=p$. A connected covering is a <regular covering map> if its <deck transformation group> acts transitively on each <fiber>; for a <path>-connected base, transitivity on the <fiber> over the chosen basepoint is equivalent.
Write $H=p_*\pi_1(\widetilde X,\widetilde b)$ and $G=\pi_1(X,b)$. A based loop represents an element of $H$ exactly when its <lift> starting at $\widetilde b$ is closed. Indeed a closed <lift> itself projects to a representative of an image element; conversely the image of a closed <path> lifts to that <path>, and the closed-endpoint property is unchanged by <based homotopy>.
Suppose the covering is regular, let $h\in H$, and let $g\in G$. <Lift> a representative of $g$ from $\widetilde b$ to an endpoint $v$. By regularity some <deck transformation> sends $\widetilde b$ to $v$. Applying it to the closed <lift> of $h$ gives a closed <lift> of $h$ at $v$, by uniqueness in the <path lifting theorem>. The concatenation of the <lift> of $g$, that closed <lift> of $h$, and the reverse <lift> of $g$ is therefore a closed <lift> of $ghg^{-1}$. Thus $gHg^{-1}\subseteq H$; applying the same argument to $g^{-1}$ gives equality. Hence \b[a regular covering has a normal image <subgroup>].
For the explicit cover, label the bottom <vertex> $B=\widetilde b$, the top $T$, the left $L$, the right $C=c$, and the rear/front equatorial vertices $U,V$, respectively. Reading the arrows gives the complete monodromy table:
|| Start
|| After $x$
|| After $y$
| $B$
| $V$
| $C$
| $V$
| $L$
| $T$
| $L$
| $T$
| $B$
| $T$
| $C$
| $U$
| $C$
| $U$
| $V$
| $U$
| $B$
| $L$
Thus the $x$ edges form the cycle $B,V,L,T,C,U,B$, and the $y$ edges form the cycle $B,C,V,T,U,L,B$. In particular every <vertex> has one incoming and one outgoing <edge> of each label, as a covering of the two-<circle> wedge requires. All word paths below are traversed from left to right.
\Image[/past-exam-of-the-topology-and-groups-course-of-the-university-of-oxford/2010-octahedral-cover.png]
{title=The two directed cycles describing the octahedral covering}
{description=Both panels use the same six named vertices. Their union is the covering graph; following x or y gives the monodromy table.}
Choose the <spanning tree> consisting of the first five $x$ edges from $B$ to $U$. The tree paths to $B,V,L,T,C,U$ have labels $1,x,x^2,x^3,x^4,x^5$. For every directed non-tree <edge> $v\xrightarrow{z}w$, the <path> from $B$ along the tree to $v$, then along the <edge>, then back along the tree from $w$, has label $t_vzt_w^{-1}$. Collapsing a <spanning tree> turns a <connected graph> into a wedge of circles, so these paths form a free basis of its <fundamental group>. There are seven non-tree edges, since $12-6+1=7$. Their projected labels give the explicit <generating set>
$$
\boxed{H=\left\langle
x^6,\ yx^{-4},\ xyx^{-3},\ x^2y,\ x^3yx^{-5},\ x^4yx^{-1},\ x^5yx^{-2}
\right\rangle.}
$$
The <covering map> induces an <injection> on fundamental groups, so these seven generators are also a free basis for the image <subgroup>. The first comes from the remaining $x$ <edge> $U\to B$; the other six come from the outgoing $y$ <edge> at each <vertex> in the table.
Now the word $h=yx^2$ lifts at $B$ around the triangle
$$
B\xrightarrow yC\xrightarrow xU\xrightarrow xB,
$$
so $h\in H$. Starting at $C$, however, its <lift> is
$$
C\xrightarrow yV\xrightarrow xL\xrightarrow xT,
$$
which is not closed. A <deck transformation> sending $B$ to $C$ would take the first closed <path> to a closed <lift> of the same word at $C$, contradicting uniqueness of lifts and the displayed nonclosed <path>. \b[No such <deck transformation> exists.] This is also the closed-<lift> obstruction expressed by the <deck transformation group as a monodromy centralizer>.
For a connected covering of a connected, locally <path>-connected, <semilocally simply connected space>, the <classification of connected covering spaces> says that the image <subgroup> is normal if and only if the covering is regular. The two-<circle> wedge and this connected finite <graph> satisfy those hypotheses. Since regularity fails, $H$ is not normal in the base <group> $\pi_1(X,b)\cong F(x,y)$. All normality comparisons here are in that base <group>.
Finally take
$$
\boxed{g=y,\qquad h=yx^2.}
$$
The <lift> of $ghg^{-1}=y^2x^2y^{-1}$ from $B$ follows
$$
B\xrightarrow yC\xrightarrow yV\xrightarrow xL\xrightarrow xT\xrightarrow{y^{-1}}V.
$$
It ends at $V\neq B$, so $ghg^{-1}\notin H$. This is the required explicit witness, and independently proves the failure of normality.
Back to article page
