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For the period lattice $\Lambda =⟨\lambda ,\mu ⟩$, define the Weierstrass elliptic function byTo prove Normal convergence of the Weierstrass elliptic-function series, let $K\subset \mathbb{C}\setminus \Lambda$ be compact and choose $R$ with $|z|\le R$ on $K$. For $|\omega |>2R$,The comparison series ${\sum }_{\omega \ne 0}|\omega {|}^{-3}$ converges by the stated lattice-sum criterion. The Weierstrass M-test gives uniform convergence on $K$; since $K$ was arbitrary, the series converges locally uniformly away from $\Lambda$. The locally uniform convergence of holomorphic functions makes its sum holomorphic there, and the displayed principal term gives a double pole at every lattice point.

The function $\wp$ is even and satisfiesThe three nonzero two-torsion points $z_1,z_2,z_3$ are distinct modulo $\Lambda$ and each equals its own negative modulo $\Lambda$. Therefore no two can have the same $\wp$-value, so $e_1,e_2,e_3$ are distinct. Equivalently, $\wp$ is a degree-two branched map and each half-lattice point already occupies its fiber with multiplicity two.

For a transitive example, take the equianharmonic latticeand define $\theta ([z])=[\omega z]$. Multiplication by $\omega$ preserves $\Lambda$, so $\theta :\mathbb{C}/\Lambda \to \mathbb{C}/\Lambda$ is a conformal equivalence. On the nonzero half-lattice points it acts bybecause ${\omega }^2=-1-\omega$. Thus it acts transitively.

Put $e=(N-1)/2$. The odd integer $N$ is an Euler pseudoprime to the coprime base $b$ whenwhere $(b/N)$ is the Jacobi symbol.

Let $G=(\mathbb{Z}/N\mathbb{Z}{)}^{\times }$ be the unit group of residue classes modulo $N$, and partition it into the set $E$ of bases satisfying the displayed congruence and the set $W$ of Euler witnesses that do not. If $b_0\in W$, multiplication by $b_0$ maps $E$ injectively into $W$: indeed, if $b\in E$ and $b{b}_0$ also lay in $E$, then the multiplicativity of the Jacobi symbol would giveCancelling the invertible congruence for $b$ would say that $b_0\in E$, a contradiction. Hence $|W|\ge |E|$, so at least half of the coprime bases are witnesses.

It remains to construct one witness for every odd composite number $N$. If $N$ is not squarefree, choose an odd prime number $p$ with $p^2\mid N$. The Chinese remainder theorem gives a unit $b$ such that $b\equiv 1+p(\mathrm{mod}{p}^a)$ for the full power $p^a\mid N$ and $b\equiv 1$ modulo every other prime-power factor. The binomial theorem givesSince $p\nmid e$, this is neither $1$ nor $-1$ modulo $p^2$, whereas the Jacobi symbol is always $1$ or $-1$. Thus $b$ is a witness.

If $N$ is squarefree and composite, choose distinct primes $p,q\mid N$. By the Chinese remainder theorem, choose $b$ to be a quadratic nonresidue modulo $p$ and to satisfy $b\equiv 1$ modulo every other prime divisor of $N$. Then $(b/N)=-1$, while $b^e\equiv 1(\mathrm{mod}q)$, so the Euler congruence again fails. This proves the existence of an Euler witness for every odd composite integer.

For subgroups $H,K\le G$ and an $H$-representation $W$, Mackey restriction formula isFrobenius reciprocity statesApplying these with $K=H$ givesThe identity double coset contributes one exactly when $W$ is irreducible, and every summand is a nonnegative integer. This proves Mackey irreducibility criterion: the induced representation is irreducible exactly when $W$ is irreducible and every nonidentity double-coset summand vanishes.

Now take $G=S_n$, $H=S_{n-1}$, and $x=(n-1n)\notin H$. Thenthe subgroup fixing both $n-1$ and $n$. Conjugation by $x$ fixes every element of this subgroup, so the two representations appearing in the corresponding Mackey summand are both ${\mathrm{Res}}_{S_{n-2}}^{S_{n-1}}W$. Their inner product is positive because this restriction is nonzero. Hence the Mackey irreducibility criterion fails whenever $W$ is irreducible. If $W$ is reducible, induction preserves its direct-sum decomposition, so the induced representation is again reducible. Therefore ${\mathrm{Ind}}_{S_{n-1}}^{S_n}W$ is never irreducible for $n\ge 2$.