Theory and applications of finite groups, by G.A. Miller, H. F. Blichfeldt [and] L. E. Dickson.

102

ABELIAN GROUPS

[Ca. IV

Sa being an arbitrary operator of G. Since
t- Itl- lsatit = sag = ti- It- scittl
it results that t and t, are commutative. On the other hand,
suppose that t is commutative with every operator in the
group of isomorphisms of G. We shall first prove that t must
transform every operator of highest order in G into a power of
itself. For, if sa is such an operator and
t - 1St = S Sa,
where s6 is not a power of s,, it is clearly possible to find an
operator tl in the group of isomorphisms of G such that t, is
commutative with sa but not with s6. As it is necessary then
that
t- It~ - ~stt=sasa, t- It- ~Stt Hss3s,,
it results that t is not commutative with every operator of the
group of isomorphisms of G unless it transforms every operator
of highest order of G into a power of itself.
We shall now show that t must transform into the same
power every operator of highest order in G, and hence it must
transform every operator of G into this power, since these operators of highest order generate G. It results from. the manner
in which the invariants of any abelian group were determined
that the group of isomorphisms of G transforms its operators
of highest order transitively. That is, the group of isomorphisms of G may be represented as a transitive substitution
group in which each letter stands for an operator of highest
order in G. If sa, so represent two operators of highest order
in G and if
t- s = sa,, t- lt = sa, y 5,
we can find an operator t2 in the group of isomorphisms of G
such that
t2- i st2 = so.
Hence it results that
t-lt2-lstot = sat, t2 —ltstls2 = s'y
That is, t, t2 are not commutative unless y= 5. This completes a proof of the theorem: A necessary and sufficient con