[proofplan] Both parts hinge on the principle that the kernel of a complex character of a finite group equals the kernel of any representation affording it: $\ker \chi := \{g \in G : \chi(g) = \chi(1)\}$ coincides with $\ker \rho := \rho^{-1}(\{I\})$. For (1), a non-identity element $g \in \ker \chi$ for some non-principal irreducible $\chi$ gives a proper normal subgroup $\ker \rho$ of order $> 1$, hence non-simplicity; conversely, a proper normal subgroup $N \neq \{1\}$ lifts a non-principal irreducible of $G/N$ to an irreducible of $G$ killing $N$. For (2), any normal $N \lhd G$ is contained in $\bigcap \ker \chi_i$ (where $\chi_i$ ranges over lifts of irreducibles of $G/N$) by the lifting construction, and the reverse containment follows because irreducible characters of $G/N$ separate non-identity elements from the identity. [/proofplan] [step:Identify $\ker \chi = \ker \rho$ for irreducible characters of a finite group] For any $g \in G$, $\rho(g) \in \operatorname{GL}(V)$ has finite order (dividing $|G|$), so $\rho(g)$ is diagonalisable over $\mathbb{C}$ with eigenvalues that are roots of unity. Let $\lambda_1, \ldots, \lambda_n$ denote these eigenvalues, where $n = \dim V$. Then $|\lambda_i| = 1$ for each $i$, and \begin{align*} \chi(g) = \operatorname{tr}(\rho(g)) = \sum_{i=1}^n \lambda_i, \qquad \chi(1) = \operatorname{tr}(I) = n. \end{align*} The triangle inequality gives $|\chi(g)| = |\sum_i \lambda_i| \leq \sum_i |\lambda_i| = n = \chi(1)$, with equality iff all $\lambda_i$ point in the same complex direction; combined with $|\lambda_i| = 1$, equality forces $\lambda_1 = \cdots = \lambda_n = \zeta$ for some unit complex number $\zeta$. In that case $\rho(g) = \zeta I$. Now suppose $g \in \ker \chi$, i.e., $\chi(g) = \chi(1) = n$. The complex number $\chi(g) = n\zeta$ is real and positive, forcing $\zeta = 1$. Hence $\rho(g) = I$ and $g \in \ker \rho$. Conversely, $g \in \ker \rho$ implies $\rho(g) = I$, hence $\chi(g) = n = \chi(1)$ and $g \in \ker \chi$. Thus $\ker \chi = \ker \rho$ as subgroups of $G$. In particular $\ker \chi \lhd G$, since $\ker \rho$ is the kernel of a homomorphism. [/step] [step:Prove the simplicity criterion in part (1)] Recall $1_G$ is the principal character, $1_G(g) = 1$ for all $g \in G$, of degree $1$. **($\Leftarrow$)** Suppose there exists an irreducible character $\chi \neq 1_G$ and an element $g \neq 1$ with $\chi(g) = \chi(1)$. By Step 1, $g \in \ker \rho$ where $\rho$ affords $\chi$. So $\ker \rho \supsetneq \{1\}$. Also $\ker \rho \neq G$: if $\ker \rho = G$ then $\rho(h) = I$ for all $h$, so $\chi(h) = \dim V$ is constant. By [Elementary Properties of Characters](/theorems/2421), constant characters are scalar multiples of $1_G$; combined with $\chi \neq 1_G$ being irreducible, $\dim V > 1$ and $\chi = (\dim V) \cdot 1_G$, but a scalar multiple of $1_G$ with multiplier $> 1$ is not irreducible (it has $1_G$ as a constituent with multiplicity $\dim V$). Hence $\ker \rho \neq G$, and $\ker \rho \lhd G$ is a proper normal subgroup of order strictly between $1$ and $|G|$. Therefore $G$ is not simple. **($\Rightarrow$)** Suppose $G$ is not simple, so there exists a normal subgroup $N \lhd G$ with $1 < |N| < |G|$. Set $\bar{G} := G/N$, a finite group of order $\geq 2$. By the [Number of Irreducible Characters](/theorems/2428), $\bar{G}$ has finitely many irreducible characters and at least one non-principal irreducible $\tilde{\chi}$ (since $\bar{G}$ has at least two conjugacy classes — the identity class and another, because $|\bar{G}| \geq 2$). By the [Lifting Lemma](/theorems/2437), the lift $\chi(g) := \tilde{\chi}(gN)$ is an irreducible character of $G$ with $N \leq \ker \chi$. We have $\chi \neq 1_G$ because $\tilde{\chi} \neq 1_{\bar{G}}$ and the lifting bijection sends $1_{\bar{G}}$ to $1_G$. Pick any $g \in N$ with $g \neq 1$, which exists since $|N| > 1$. Then $g \in N \leq \ker \chi$, so $\chi(g) = \chi(1)$, completing the implication. [guided] The criterion is that $G$ has a proper normal subgroup of order $> 1$ precisely when some non-principal irreducible has a kernel of order $> 1$. The proof is two implications, both routed through Step 1's identification $\ker \chi = \ker \rho$. For ($\Leftarrow$): given $\chi \neq 1_G$ and $g \neq 1$ with $\chi(g) = \chi(1)$, Step 1 gives $g \in \ker \rho$ for any $\rho$ affording $\chi$. So $\ker \rho \supsetneq \{1\}$. We must check it is proper. If $\ker \rho = G$, the representation $\rho$ is the constant identity, $\chi$ is a positive constant, and irreducibility forces $\chi = 1_G$ (a constant character of degree $> 1$ decomposes as a multiple of $1_G$, hence fails irreducibility) — contradicting $\chi \neq 1_G$. So $\ker \rho \lhd G$ is a proper normal subgroup with order strictly between $1$ and $|G|$. For ($\Rightarrow$): suppose $1 < N \lhd G$ is proper. Apply representation theory to $\bar{G} = G/N$. As a finite group of order $\geq 2$, $\bar{G}$ has at least two conjugacy classes (the class of the identity and the class of any non-identity element), so by the [Number of Irreducible Characters](/theorems/2428) at least two irreducible characters: $1_{\bar{G}}$ and at least one more, $\tilde{\chi} \neq 1_{\bar{G}}$. Lift $\tilde{\chi}$ to $\chi$ on $G$ via the [Lifting Lemma](/theorems/2437); the lift is irreducible, distinct from $1_G$, and satisfies $\chi(k) = \tilde{\chi}(kN) = \tilde{\chi}(N) = \chi(1)$ for every $k \in N$. Picking any non-identity $g \in N$ gives the desired pair. [/guided] [/step] [step:Prove $N \leq \bigcap_i \ker \chi_i$ in part (2) by lifting characters of $G/N$] Let $N \lhd G$. Let $\tilde{\chi}_1, \ldots, \tilde{\chi}_\ell$ enumerate the irreducible characters of $G/N$ (where $\ell$ is the number of conjugacy classes of $G/N$), and let $\chi_1, \ldots, \chi_\ell$ be their lifts to $G$ via the [Lifting Lemma](/theorems/2437): $\chi_i(g) = \tilde{\chi}_i(gN)$. By the lemma, each $\chi_i$ is an irreducible character of $G$. For any $k \in N$ and any $i$, \begin{align*} \chi_i(k) = \tilde{\chi}_i(kN) = \tilde{\chi}_i(N) = \tilde{\chi}_i(1_{G/N}) = \chi_i(1), \end{align*} using $kN = N$ for $k \in N$. Hence $k \in \ker \chi_i$ for every $i$, giving $N \leq \bigcap_{i=1}^\ell \ker \chi_i$. In particular, the index set $I := \{i_1, \ldots, i_\ell\}$ corresponds to those irreducible characters of $G$ obtained as lifts from $G/N$, i.e., those satisfying $N \leq \ker \chi$. [/step] [step:Prove the reverse inclusion $\bigcap_i \ker \chi_i \leq N$ via separation by characters of $G/N$] We show: for $g \in G \setminus N$, there exists some $\chi_i$ in our list with $\chi_i(g) \neq \chi_i(1)$, hence $g \notin \ker \chi_i$ and so $g \notin \bigcap_i \ker \chi_i$. Fix $g \in G \setminus N$. Then $gN \neq N$ in $G/N$, so $gN$ is a non-identity element of $G/N$. By the [Number of Irreducible Characters](/theorems/2428) applied to $G/N$, the irreducible characters $\tilde{\chi}_1, \ldots, \tilde{\chi}_\ell$ form an orthonormal basis of the space of class functions $\mathcal{C}(G/N)$, of dimension equal to the number of conjugacy classes of $G/N$. The indicator function of the identity class $\{1_{G/N}\}$ is a non-zero class function and so has a non-zero expansion in this basis; equivalently, the irreducible characters separate elements of $G/N$ in the following sense. [claim:The irreducible characters of any finite group separate the identity from any non-identity element] [proof] Let $H$ be a finite group and let $h \in H$ with $h \neq 1_H$. Suppose for contradiction that $\psi(h) = \psi(1_H)$ for every irreducible character $\psi$ of $H$. Let $\rho_{\mathrm{reg}}: H \to \operatorname{GL}(\mathbb{C}[H])$ be the regular representation, with character $\pi_{\mathrm{reg}}$. By the decomposition of the regular representation into irreducibles (a consequence of [Maschke's Theorem](/theorems/2409) and orthogonality), $\pi_{\mathrm{reg}} = \sum_\psi \psi(1_H)\, \psi$, where the sum runs over irreducible characters $\psi$ of $H$. By hypothesis, each $\psi(h) = \psi(1_H)$, so \begin{align*} \pi_{\mathrm{reg}}(h) = \sum_\psi \psi(1_H) \psi(h) = \sum_\psi \psi(1_H)^2 = \pi_{\mathrm{reg}}(1_H) = |H|. \end{align*} But by direct computation, $\pi_{\mathrm{reg}}(h) = |\{x \in H : hx = x\}|$, which equals $|H|$ if $h = 1_H$ and $0$ otherwise. So $\pi_{\mathrm{reg}}(h) = 0$, contradicting $\pi_{\mathrm{reg}}(h) = |H|$ (and $|H| \geq 1$). Hence some $\psi$ satisfies $\psi(h) \neq \psi(1_H)$. [/proof] [/claim] Apply the claim to $H = G/N$ and $h = gN \neq 1_{G/N}$: there exists an index $i$ with $\tilde{\chi}_i(gN) \neq \tilde{\chi}_i(1_{G/N}) = \tilde{\chi}_i(N)$. By the lift formula, \begin{align*} \chi_i(g) = \tilde{\chi}_i(gN) \neq \tilde{\chi}_i(N) = \chi_i(1), \end{align*} so $g \notin \ker \chi_i$ and hence $g \notin \bigcap_j \ker \chi_j$. Combining with Step 3, $\bigcap_{i=1}^\ell \ker \chi_i = N$. [guided] The two-step strategy is the standard pattern for "$N$ is determined by the characters of $G/N$": (a) Every lift $\chi$ of an irreducible $\tilde{\chi}$ of $G/N$ satisfies $N \leq \ker \chi$. This is automatic: $\chi(k) = \tilde{\chi}(kN) = \tilde{\chi}(N) = \chi(1)$ for $k \in N$. (b) Conversely, any $g \notin N$ is detected by some lift. The reason is that the irreducible characters of $G/N$ separate $1_{G/N}$ from other elements of $G/N$ — a fact that is a corollary of completeness of the irreducible characters (or equivalently of the regular representation decomposition). Lifting back, the corresponding lift $\chi_i$ has $\chi_i(g) \neq \chi_i(1)$, so $g \notin \ker \chi_i$. The separation claim is proved by computing $\pi_{\mathrm{reg}}(h)$ in two ways: via the basis decomposition $\pi_{\mathrm{reg}} = \sum_\psi \psi(1) \psi$, and directly as $\pi_{\mathrm{reg}}(h) = |\{x : hx = x\}|$. The direct value is $0$ for $h \neq 1$ (no element of $H$ is fixed by left-multiplication by a non-identity element). The basis decomposition value is $|H|$ if all $\psi(h) = \psi(1)$. These contradict, so some $\psi(h) \neq \psi(1)$. This is the substance of "characters separate elements" — the version for the identity vs. non-identity is enough for our purposes. The full version (any two distinct elements separated by some character) follows by applying this argument to $h = ab^{-1}$ for $a \neq b$. The conclusion $N = \bigcap_{N \leq \ker \chi_i} \ker \chi_i$ — every normal subgroup is the intersection of kernels of irreducible characters containing it — is the converse of the elementary fact that any kernel is normal. So the kernels of irreducible characters generate the lattice of normal subgroups under intersection. [/guided] [/step]