# Normality is upper join-closed

This article gives the statement, and possibly proof, of a subgroup property (i.e., normal subgroup) satisfying a subgroup metaproperty (i.e., upper join-closed subgroup property)
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## Statement

### Statement with symbols

Suppose $H$ is a subgroup of $G$, $I$ is a nonempty indexing set, and $K_i, i \in I$ are subgroups of $G$ containing $H$, such that $H \triangleleft K_i$ (i.e., $H$ is a normal subgroup of $K_i$) for each $i \in I$. Then, $H$ is normal in the join of the $K_i$s.

## Related facts

### Related facts about upper join-closedness

The fact about normality generalizes to the following:

Left-inner right-monoidal implies upper join-closed: A subgroup property that has a function restriction expression with the left property being inner automorphisms and the right property being monoidal (closed under composition) is upper join-closed.

Other manifestations of the general fact include:

Here are some related properties that are not upper join-closed:

### Analogues and breakdowns of analogues in other algebraic structures

• Ideal property is upper join-closed for Lie rings: If $I$ is a subring of a Lie ring $L$ such that $I$ is an ideal in two subrings $A_j\le L$, where $j \in J$, an indexing set, then $I$ is also an ideal in the Lie subring generated by all the $A_j$s.
• Ideal property is not upper join-closed for alternating rings
• Normality is not upper join-closed for algebra loops

## Proof

Given: A group $G$, a subgroup $H$, a nonempty indexing set $I$, and a collection of subgroups $K_i, i \in I$, such that $H$ is normal in $K_i$ for each $i \in I$.

To prove: $H$ is normal in the join of the $K_i$s.

Proof: Let $K$ be the join of the $K_i$s. For $g \in K$, we can write: $g = g_1g_2g_3\dots g_n$

where $g_j \in K_{i_j}$ for some index element $i_j$. Thus, if $c_g$ denotes conjugation by $g$, we have: $c_g = c_{g_1} \circ c_{g_2} \circ \dots \circ c_{g_n}$

Now, since $H$ is normal in $K_{i_j}$, each $c_{g_j}$ acts as an automorphism of $H$. Thus, their composite, namely $c_g$, is also an automorphism of $H$. In other words, $c_g(H) = H$ for every $g \in K$, showing that $H$ is normal in $K$.