# Transitive normality is not finite-join-closed

This article gives the statement, and possibly proof, of a subgroup property (i.e., transitively normal subgroup) not satisfying a subgroup metaproperty (i.e., finite-join-closed subgroup property).This also implies that it does not satisfy the subgroup metaproperty/metaproperties: Strongly finite-join-closed subgroup property (?), .
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## Statement

### Verbal statement

The join of two transitively normal subgroups of a group need not be transitively normal.

### Statement with symbols

Suppose $G$ is a group and $H, K$ are two transitively normal subgroups of $G$. Then, the join $\langle H, K \rangle$, which in this case equals the product $HK$, need not be a transitively normal subgroup.

## Related facts

### Related properties that are join-closed

• Normality is strongly join-closed: In particular, this implies that any join of transitively normal subgroups is a normal subgroup, even though it need not be transitively normal.

## Proof

Further information: symmetric group:S3

Let $A$ be the cyclic group of order three. Let $B$ be the symmetric group of degree three, and $C$ be the subgroup of order three in $B$. Define: $G = A \times B, \qquad H = A \times 1, \qquad K = 1 \times C$.

Then, we have:

• $H$ and $K$ are both normal subgroups of prime order. In particular, the only normal subgroups they have are the whole group and the trivial subgroup, both of which are normal in $G$. Thus, $H$ and $K$ are both transitively normal.
• $HK$ is not transitively normal: $HK = A \times C$. $HK$ is elementary abelian of order nine. Pick an isomorphism $\sigma:A \to C$ and consider the subgroup $L$ of $HK$ given by $L = \{ (a, \sigma(a) \mid a \in A \}$. Then, $L$ is normal in $HK$, since $HK$ is abelian. However, $L$ is not normal in $G$, because conjugation by an element in $1 \times (B \setminus C)$ sends $(a,\sigma(a))$ to $(a,\sigma(a)^2)$, which is not in $L$. Thus, $HK$ is not transitively normal.

Note that in this example, $H$ is the center of $G$ and is also a direct factor of $G$. This shows that taking the join of a transitively normal subgroup with the center or with a direct factor does not guarantee a transitively normal subgroup.