Re: Nilpotent groups
- From: magidin@xxxxxxxxxxxxxxxxx (Arturo Magidin)
- Date: Tue, 16 May 2006 15:50:26 +0000 (UTC)
In article <1147793247.143968.151010@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Richard Hayden <r.hayden@xxxxxxxxx> wrote:
Hi,
I have a proof that if G is nilpotent, every proper subgroup is
properly contained in its normaliser. I can't understand the last step
in the proof and I'd be grateful if someone could clarify it for me
(Z(G) is the centre of G and N_G(H) is the normaliser of H in G):
Assume H < G. Clearly zHz^(-1) = H for all z in Z(G), so N_G(H) >=
Z(G)H. We have the result unless Z(G) <= H. So we assume that. If Z(G)
<= H, then G/Z(G) > H/Z(G). So N_G(H)/Z(G) > H/Z(G) by induction on |G|
(*this sentence is the step I do not understand*). Hence the result.
Well, it is more or less a standard argument, and (dis)organized in a
more or less standard way. Let me rephrase it so that it looks exactly
like an induction. The thing is that in this case, the inductive step
will be established either directly, or else by appealing to a smaller
case, rather than inductions you may be more familiar with (which
always use the smaller case; but for an example like this, see the
proof of the Fundamental Theorem of Arithmetic, where in the inductive
step you argue that if n is prime you are done, and if not, then...)
If the order of |G| is 2, 3, 4, or 5, then the result is true because
the group G will be abelian, hence the normalizer of any proper
subgroup is G, which properly contains H.
Assume the order of G is n>5, and that the result holds for any
nilpotent group K of order |K|=k<n. Let H be a proper subgroup of G.
Since G is nilpotent, we know the center is nontrivial. We have two
cases: either H contains Z(G), or H does not contain all of Z(G).
If H does not contain the center, then HZ(G) is a subgroup of G that
properly contains H and that normalizes H, hence HZ(G) is contained in
N_G(H), which must properly contain H (since HZ(G) properly contains
H); so the result is proven in this case.
If H does contain the center, then since Z(G) is normal in G, we can
consider the subgroup H/Z(G) of G/Z(G) = K. Since Z(G) is nontrivial,
we know that the order of K is strictly smaller than the order of G,
hence by induction we know that the normalizer in K of H/Z(G) properly
contains H/Z(G). Let N be the subgroup of G that corresponds to the
normalizer of H/Z(G) in K; by the isomorphism theorems, this properly
contains H. We claim that N normalizes H. Indeed, let n in N; since
the image of n normalizes H/Z(G), that means that the image of
nHn^{-1} must be H/Z(G). By the isomorphism theorems, the only
subgroup that contains Z(G) and corresponds to H/Z(G) is H, so
nHn^{-1}=H. Thus, n normalizes H.
Since N properly contains H, and normalizes H, that shows that, in the
case where H does not contain Z(G), H is properly contained in the
normalizer as well.
This proves the induction step, and so the theorem. QED
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
magidin@xxxxxxxxxxxxxxxxx
.
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