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> [!proposition] Proposition. ([[normal subgroups from character table]])
> Using [[normal subgroup iff intersection of irrep character kernels]] and recalling [[normal subgroup iff union of conjugacy classes]], we can obtain all [[normal subgroup|normal subgroups]] of a finite [[group]] $G$ by taking all possible intersections of kernels of the (irreducible) characters.
> [!basicexample]
> First consider by example $Q_{8}$:
>
| $Q_{8}$ | $[\mathbf{1}]$ | $[-\mathbf{1}]$ | $[\mathbf{i}]$ | $[\mathbf{j}]$ | $[\mathbf{k}]$ |
|----------------------------|----------------|-----------------|----------------|----------------|----------------|
| $\chi_{\mathbb{1}}$ | 1 | 1 | 1 | 1 | 1 |
| $\chi_{(-1,1)}$ | 1 | 1 | -1 | 1 | -1 |
| $\chi_{(1,-1)}$ | 1 | 1 | 1 | -1 | -1 |
| $\chi_{(-1,-1)}$ | 1 | 1 | -1 | -1 | 1 |
| $\chi_5$ | 2 | -2 | 0 | 0 | 0 |
\
Recall that [[quaternion group is nonabelian, but all its subgroups are normal]]. It is enough to then check all possible intersections of [[irreducible group representation|irreducible]] kernels:
>- $\ker \chi_{\mathbb{1}} \implies Q_{8} \trianglelefteq Q_{8}$
>- $\ker \chi_{(-1,1)} \implies$ $\langle \v j \rangle \trianglelefteq Q_{8}$
>- $\ker \chi_{(1,-1)} \implies \langle \v i \rangle \trianglelefteq Q_{8}$
>- $\ker \chi_{(-1,-1)} \implies \langle \v k \rangle \trianglelefteq Q_{8}$
>- $\ker \chi_{(-1,-1)} \cap \ker \chi_{(1,-1)} \implies \{ \v 1, - \b 1 \} \trianglelefteq Q_{8}$
>- $\ker \chi_{5} \implies$ $(\v 1) \trianglelefteq Q_{8}$
>
The rest of the intersections give duplicates of what we already have obtained; the [[lattice (groups)]] of [[subgroup|subgroups]] derived [[quaternion group|here]] confirms this.
> [!proposition] Corollary. (Character table test for simplicity)
> $G$ has a proper normal nontrivial subgroup if and only if a nontrival irrep character has nontrivial kernel. A proof that does not use [[normal subgroup iff intersection of irrep character kernels]] is here [[character table test for simplicity]].
> [!proof]- Proof. ([[normal subgroups from character table]])
> Pretty much immediate from [[normal subgroup iff intersection of irrep character kernels]]: that tells us that the normal subgroups of $G$ are obtained exactly by taking all possible intersections of irrep kernel characters, which we have access to in the [[character table]].
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#### References
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