Zassenhaus isomorphism theorem
From Groupprops
This article describes a fact or result that is not basic but it still well-established and standard. The fact may involve terms that are themselves non-basic
View other semi-basic facts in group theory
VIEW FACTS USING THIS: directly | directly or indirectly, upto two steps | directly or indirectly, upto three steps|
VIEW: Survey articles about this
This article is about an isomorphism theorem in group theory.
View a complete list of isomorphism theorems| Read a survey article about the isomorphism theorems
Contents
Statement
Suppose are subgroups of a group
such that
is a normal subgroup of
and
is a normal subgroup of
. Then,
is normal in
and
is normal in
, and we have an isomorphism:
.
Facts used
- Normality satisfies transfer condition: If
is normal in
and
is a subgroup of
,
is normal in
.
- Modular property of groups: If
, then
and
.
- Join lemma for normal subgroup of subgroup with normal subgroup of whole group: If
and
is normal in
, then the subgroup
is normal in the subgroup
.
- Second isomorphism theorem: If
are subgroups of a group
such that
is normal in
, then
is isomorphic to
.
Proof
Hands-on proof
Given: are subgroups of a group
such that
is a normal subgroup of
and
is a normal subgroup of
.
To prove: is normal in
and
, and we have an isomorphism:
.
Proof:
- (Given data used:
; Fact used: fact (1)):
is normal in
: This follows from fact (1), setting
.
- (Given data used:
):
and
normalize
: Since
is normal in
, the normalizer of
contains
, hence it also contains
.
-
and
are subgroups: This follows directly from step (2).
-
is normal in
: By step (2),
is normal in
, and by step (1),
is normal in
. Thus, by fact (3), we obtain that
is normal in
. (Here, we set
).
- (Facts used: fact (4)): Setting
and
, we observe that by step (4),
is normal in
, so applying fact (4) yields:
-
.
-
- (Facts used: fact (2)): Applying fact (2) with
yields
. (Note that both are equal because of the fact that
, being normal in
, permutes with
. We thus have:
-
.
-
-
is normal in
, both are subgroups, and we have (by analogous reasoning to steps (1) - (6)):
-
.
-
- The right sides for steps (6) and (7) are equal, hence the left sides are isomorphic, completing the proof.