1
$\begingroup$

I have to answer the following two questions:

  1. Using Zorn's Lemma, one could try to give a "proof" of the following statement: Every subgroup $H$ of a group $G$ such that $H \neq G$ is contained in a maximal subgroup of $G$. Explain why this argument does not work.
  2. Is it true that any group has a maximal normal subgroup?

For #1, I showed just prior to this that the group $\mathbb{Q}$ has no maximal subgroup. Is that the reason why the argument does not work - because not every group has a maximal subgroup? Or is there more to it than this?

For #2, again, simple groups have no proper normal subgroups, and so therefore, can have no maximal normal subgroup (since, in order for a subgroup to be maximal, it first and foremost has to be proper). Again, is this the reason why the answer is no, or is there more to it than this?

In case I am missing the real reasons for these, a brief outline of how I should approach 1 and 2 would be appreciated. Thank you.

$\endgroup$
2
$\begingroup$

Keep in mind the intent of exercise (1) is to point out that the precondition for applying Zorn's lemma - the hypothesis that all chains of subgroups have upper bounds - does not hold true. Your example with $\mathbb{Q}$ works fine for that.

With exercise (2), the term maximal normal subgroup allows for the trivial group to be an example, which means even simple groups have maximal normal subgroups. Just use $\mathbb{Q}$ as a counterexample again - it's abelian so all subgroups are normal, and a "maximal normal subgroup" would just be a maximal subgroup then.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy