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When $[G:H]=2$, there are exactly two left cosets and two right cosets. For $g\notin H$, both $gH$ and $Hg$ are the complement $G\setminus H$; for $g\in H$, both equal $H$. This is the index-two subgroup is normal argument: $gH=Hg$ for every $g$, which is equivalent to $gH{g}^{-1}=H$. Therefore every subgroup of index two is normal.

In the dihedral group $D_{2n}=⟨r,s⟩$, the rotation cyclic subgroup $⟨r⟩$ has order $n$ and index two, so it is a normal subgroup. It is nontrivial and proper for the usual dihedral convention $n\ge 2$ (polygon symmetries have $n\ge 3$). If the degenerate convention $D_2=C_2$ were allowed, the claimed nontrivial proper subgroup would not exist; the rotation argument uses $n>1$.

For any $k\ge 3$, let $G=S_k$ and take the stabilizer subgroup $H$ of the letter $1$. Its elements freely permute the other $k-1$ letters, soConjugation by $(12)$ sends $H$ to the stabilizer of $2$, which is different: for instance $(23)$ belongs to $H$ but does not fix $2$. Thusgives a nonnormal subgroup of every prescribed index $k\ge 3$.

Because $L/K$ is a finite Galois extension,

Suppose the irreducible quadratic $f$ became reducible over $L$. It would then have a root $\alpha \in L$, and $K(\alpha )/K$ would be a quadratic intermediate extension. By the Fundamental theorem of Galois theory, $\mathrm{Gal}(L/K(\alpha ))$ would be a subgroup of index two in $A_{10}$ and hence a normal subgroup by the index-two subgroup is normal result. But the Simplicity of alternating groups says that $A_{10}$ is simple, while this subgroup would be proper and nontrivial. This contradiction proves that $f$ remains irreducible in $L[x]$.

Choose a path $\gamma$ in $X$ from $x_0$ to $x_1$. For every $\hat{x}\in p^{-1}(x_0)$, the path lifting theorem gives a unique lift of $\gamma$ starting at $\hat{x}$. Send $\hat{x}$ to the endpoint of this lift. Lifting the reversed path gives the inverse map, so this is the fibre bijection by path liftingIn particular, all fibres have the same cardinality.

A connected covering is a normal covering map when its deck transformations act transitively on a fibre. Equivalently, for a choice of ${\hat{x}}_0$ above $x_0$,The relevant lifting criterion for a covering space says that a based map $f:(Y,y_0)\to (X,x_0)$ lifts through $p$ exactly whenFor a universal covering, $\hat{X}$ is simply connected, so the displayed covering subgroup is trivial and hence normal. Thus a universal covering map is normal.

Now consider connected finite covers of the closed orientable surface ${\Sigma }_g$. By the classification of connected covering spaces, degree-$n$ connected covers correspond to index-$n$ subgroups of the fundamental group of a closed orientable surface, and normal covers correspond to normal subgroups.

The cases in which normality is forced are:

It remains to show that these are the only forced cases. Let $g\ge 2$ and $n\ge 3$. WriteIn the symmetric group $S_n$, putDefine a homomorphism byand send all remaining generators to the identity. The relation is respected becauseThe cycle $\sigma$ and transposition $\tau$ generate $S_n$, so the homomorphism is surjective.

LetThe natural action of $S_n$ is transitive, so $H$ has index $n$. Its image is the point stabilizer $S_{n-1}$, which is not normal in $S_n$ for $n\ge 3$; hence $H$ is not normal. The connected covering corresponding to $H$ is therefore an explicit degree-$n$ nonnormal cover. This is the nonnormal finite cover of a higher-genus orientable surface.

Consequently, among connected covers that exist, normality is forced exactly whenwith the qualification that $g=0$ permits only $n=1$, as summarized by the forced normality of finite connected covers of orientable surfaces.

A normal subgroup $H$ satisfies $gH{g}^{-1}=H$ for every $g\in G$, equivalently $gH=Hg$. A left coset is $gH=\{gh:h\in H\}$ and a right coset is $Hg=\{hg:h\in H\}$.

If $|G|=2|H|$, then for $g\in H$ both cosets equal $H$. For $g\notin H$, neither coset intersects $H$: an intersection would imply $g\in H$. Both have $|H|$ elements, because multiplication by $g$ is a bijection. They therefore both equal $G\setminus H$, proving $gH=Hg$ in this case too. This establishes that an index-two subgroup is normal.

For Lagrange's theorem, left cosets partition $G$. Indeed if $g_1{h}_1=g_2{h}_2$, then $g_{1}H=g_{2}H$ by multiplying by elements of $H$; otherwise the cosets are disjoint. Every coset has $|H|$ elements, so if there are $[G:H]$ cosets,The group order of $H$ thus divides the group order of $G$. Normality is not needed for this divisibility statement.