I have this classic exercise to prove:

If $ \{ N_i : i \in I \} $ is a family of normal subgroups of a group $ G $, then $ \bigcap_{i \in I} N_i $ is a normal subgroup of $ G $ too.

I am searching for a constructive reasoning.

My solution is like as follows:

Let's explicite the set of indices: $ I_n = \{ 1,2,...,n \} $. And let's give a name to the set mentioned above: $ F_i = \{ N_i : \forall i \in I_n, N_i < G \} $. The crucial difference is that: we will suppose that elements of our set $ F $ are not normal subgroups of G. Because we know that this enunciation is a valid property when $ N_i $ are ordinary subgroups.

Then define a function $ f : F_i \to \wp (G) $ such that, for $ i = 1 $ we have $ f(1) = N_1 $, and for $ i \ge 2 $, we have $ f(i) = N_i \bigcap f(N_{i-1}) $.

We can easily see that: $ \forall i \in I_n, f(i) \subset \wp (G) $, even better: $ f(i) < G $. As we can check: $ Dom(f) = F_i $ is a group on its own under the usual law of intersection; and the same for $ Im(f) = \wp (G) $.

So the tactic is to check if this function $ f $ is a homomorphism of groups? If it is so, then $ f $ preserves algebraic properties, and, as we started with the known conclusion (the case where sets are not normal subgroups) and constructed our correspondence, we can conclude that the weaker property will be inherited.

My need for help begins here: I don't know if $ f $ is well defined, AND, I couldn't prove that my function $ f $ is indeed a homomorphism.

Thank you all in advance for any advice and correction.

  • $\begingroup$ What is $\wp (G)$? $\endgroup$
    – user169852
    Dec 22, 2018 at 0:54
  • $\begingroup$ It is the Power Set of G. I couldn't find the appropriate symbol on Jax. $\endgroup$ Dec 22, 2018 at 0:55
  • $\begingroup$ And what is the assumed group operation on $\wp(G)$? $\endgroup$
    – user169852
    Dec 22, 2018 at 0:56
  • 4
    $\begingroup$ You're overcomplicating this. You also make a mistake in the very first sentence: the indices needn't be countable. If each $N_i$ is normal, it follows that for each $g \in G$ and each $n_i \in N_i$, we have $gn_ig^{-1} \in N_i$. Hence if $n^{*} \in \bigcap N_i$ and $g \in G$, what can we say about $gn^{*}g^{-1}$? $\endgroup$ Dec 22, 2018 at 0:57
  • $\begingroup$ Usual intersection. We need not Subsets of G be Subgroups. We need closure, associativity, etc. for Power Set of G to be a Group only. $\endgroup$ Dec 22, 2018 at 0:58

1 Answer 1


Much simpler: take $\;x\in\bigcap_{i\in I} N_i\;$ , then for any $\;g\in G\;$ :

$$\forall\,i\in I\;,\;\;x^g:=gxg^{-1}\in N_i\implies x^g\in\bigcap _{i\in I}N_i\;\ldots$$

Observe that in the above it isn't assumed the index set $\;I\;$ is countable: that doesn't affect.


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