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Take $X=S^1\vee S^1$, the wedge of two circles, and label its oriented loops $a$ and $b$. For a fixed $n\ge 3$, letbe permutations of the fibre $\{1,\dots ,n\}$. Construct a labelled graph ${\hat{X}}_n$ with these $n$ vertices, an oriented $a$-edge from $i$ to $\alpha (i)$, and an oriented $b$-edge from $i$ to $\beta (i)$. Map every vertex to the wedge point and each labelled edge homeomorphically onto its corresponding loop. Every vertex has exactly one incoming and one outgoing edge of each label, so this is an $n$-sheeted permutation covering of a wedge of circlesIt is connected because the $n$-cycle $\alpha$ acts transitively on the fibre.

A deck transformation induces a permutation $\gamma$ of the vertices that commutes with both monodromy permutations $\alpha$ and $\beta$. Conversely, such a permutation determines a deck transformation, by the deck transformation group as a monodromy centralizer. The permutations $\alpha$ and $\beta$ generate $S_n$: the conjugates ${\alpha }^j\beta {\alpha }^{-j}$ include the adjacent transpositions, which generate the symmetric group. Hence $\gamma$ centralizes $S_n$ and lies in its center of a group. For $n\ge 3$, the center of $S_n$ is trivial, so $\gamma =1$. A deck transformation fixing one point is the identity by part (a), and therefore

No. The monodromy action of a covering space for a two-sheeted covering takes values in $S_2$. Its nonidentity transposition commutes with every subgroup of $S_2$, so the deck transformation group as a monodromy centralizer always contains that transposition. Equivalently, every double cover has a deck involution of a double covering that exchanges the two points in each fibre. Thus a two-sheeted cover cannot have trivial deck group.

A deck transformation of a covering $p:\tilde{X}\to X$ is a homeomorphism $d:\tilde{X}\to \tilde{X}$ 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 $H=p_{\ast }{\pi }_1(\tilde{X},\tilde{b})$ and $G={\pi }_1(X,b)$. A based loop represents an element of $H$ exactly when its lift starting at $\tilde{b}$ 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 $\tilde{b}$ to an endpoint $v$. By regularity some deck transformation sends $\tilde{b}$ 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 $gh{g}^{-1}$. Thus $gH{g}^{-1}\subseteq H$; applying the same argument to $g^{-1}$ gives equality. Hence a regular covering has a normal image subgroup.

For the explicit cover, label the bottom vertex $B=\tilde{b}$, 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:

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.

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 $v\stackrel{z}{\to }w$, the path from $B$ along the tree to $v$, then along the edge, then back along the tree from $w$, has label $t_{v}z{t}_w^{-1}$. 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 setThe 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 $h=y{x}^2$ lifts at $B$ around the triangleso $h\in H$. Starting at $C$, however, its lift iswhich 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. 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 takeThe lift of $gh{g}^{-1}=y^2{x}^2{y}^{-1}$ from $B$ followsIt ends at $V\ne B$, so $gh{g}^{-1}\notin H$. This is the required explicit witness, and independently proves the failure of normality.