How to prove $Z(G)$ is not a maximal subgroup of $G$?
So I used the fact that $Z(G)$ is a subset of the $C(a)$ which is a subgroup of some group $G$. Is that sufficient? If not can you provide a proof?
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Hint Assume that $Z(G)\ne G$ (why?)
Then there exists $a\in G$ such that $a$ is not in $Z(G)$
You know that $Z(G)\le C(a) \le G$
But here, $Z(G)< C(a) < G$ (why < ?)
And with this you're done showing $Z(G)$ is not maximal in $G$
Swapnil Tripathi
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I hope you know the notations $<$ and $\le$. The former is for proper subgroup – Swapnil Tripathi Nov 16 '14 at 00:09
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I am lost as to why you assume Z(G) is not G. – fr56 Nov 16 '14 at 00:20
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1Does it follow that since Z(G) = intersection C(a) for a in G that Z(G)<C(a)<G? – fr56 Nov 16 '14 at 00:28
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What is the definition of maximal subgroup? – Swapnil Tripathi Nov 16 '14 at 00:33
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A proper subgroup H of a group G is called maiximal if there is no subgroup K such that H<K<G. @Swapnil Tripathi – fr56 Nov 16 '14 at 00:35
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1For $Z(G)<C(a)$ there is a very obvious element in $C(a)$ that is not in $Z(G)$. What is it? For $C(a)<G$, suppose $C(a)=G$ what can you say about $a$? What contradiction do you reach. – Swapnil Tripathi Nov 16 '14 at 00:38
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Yes, since you can talk about maximality only when $H$ (in this case $Z(G)$) is proper. That is, $H\ne G$ – Swapnil Tripathi Nov 16 '14 at 00:40
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The element a is in C(a) but not in Z(G)? @Swapnil Tripathi – fr56 Nov 16 '14 at 00:46
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Yes. Absolutely right. What about the contradiction when $C(a)=G$? – Swapnil Tripathi Nov 16 '14 at 00:47
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I am a little stuck on that part, can you guide me? @Swapnil Tripathi – fr56 Nov 16 '14 at 00:51
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If $C(a)=G$. Then, $a$ commutes with all elements of $G$. Contradiction? – Swapnil Tripathi Nov 16 '14 at 00:53
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Oh man! I was just about to comment that, I lack conviction. Thanks. One other thing: why do we assume that Z(G)=!G? – fr56 Nov 16 '14 at 00:55
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Reread all the comments. You'll find it. – Swapnil Tripathi Nov 16 '14 at 00:56
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Got it! Thank you for all your help! – fr56 Nov 16 '14 at 00:58
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Hint If $Z(G)$ is maximal, then $G/Z(G)$ doesn't have any non-trivial subgroup.
This implies that $G/Z(G)$ has a prime order and hence is cyclic.
Now prove that this implies that every two elements of $G$ commute, which leads to $Z(G)=G$, contradiction.
N. S.
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Why does Z(G) maximal imply G/Z(G) does not have any non-trivial subgroup? – fr56 Nov 15 '14 at 23:53
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The subgroup correspondence theorem states that subgroups of $G/N$ correspond to subgroups of $G$ containing $N$, where $N$ is normal in $G$. Thus, if $G/Z(G)$ contains a non-trivial subgroup, then $G$ has a subgroup containing $Z(G)$, so $Z(G)$ is not maximal. – William Stagner Nov 16 '14 at 00:03
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@WilliamStagner I have never seen that theorem. What other way can I go about it? – fr56 Nov 16 '14 at 00:09
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1@fr, if you haven't seen the correspondence then google it. It is one of the most basic, important theorem in basic group theory. – Timbuc Nov 16 '14 at 00:33