Let $ G $ is a finite group. Permutizer subgroup $ H $ of $ G $ is $ P_{G}(H) = \langle g \in G \ \vert \ \langle g \rangle H = H \langle g \rangle \rangle \ $. If for any proper subgroup $ H $ of $ G $, $ H < P_{G}(H) $ then we say $ G $ satisfies the permutizer condition. $ G $ satisfies the maxiaml permutizer condition if for any maximal subgroup $ M $ of $ G $, $ P_{G}(M) = G $.
Now I want show $ D_{8} = \langle a , b \ \vert a^{4} = b^{2} = 1 \ , \ bab^{-1} = a^{-1} \rangle $ satisfies the permutizer condition and maximal permutizer condition or no?
We know $ D_{8} $ has $ 3 $ subgroups of order $ 4 $ and $ 5 $ subgroups of order $ 2 $. Subgroups of order $ 4 $ are normal in $ D_{8} $ and subgroups of order $ 2 $ are isomorphic to $ Z_{2} $ and so cyclic, Hence $ G $ satisfies the permutizer and maximal permutizer condition.
Is this true or no?