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Maximal subgroups and generators

Show that

x \in G

lies in the intersection of all maximal subgroups of G if and only if it has the following property: if

X \subseteq G

contains

x

and generates

G

, then

X\setminus\{x\}

generates

G

.

Maximal subgroups are defined to be the largest possible subgroups that are generated by a certain set.

So the first part of the exercise is alright. If

x \in \bigcap(M_i)

, where

M_i

is the set of all maximal subgroups, then the cyclic subgroup generated by x is obviously also in the group generated by

X\setminus\{x\}

, so

G = \langle {X} \rangle = \langle{X}\setminus\{x\}\rangle

.

For the converse however, I have got into a little bit of a problem. Take for example the Klein 4-group.

\langle V_4 \setminus \{a\}\rangle = \langle 1, b, c \enspace | \enspace b^2 = c^2 = 1, bc = cb = a \rangle

, clearly generates

\{1,a,b,c\}

, however

a \notin \langle b \rangle, \langle c \rangle

, so the converse is not always true?

Where have I gone wrong?

(edited 9 years ago)

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