# Subnormality satisfies transfer condition

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This article gives the statement, and possibly proof, of a subgroup property (i.e., subnormal subgroup) satisfying a subgroup metaproperty (i.e., transfer condition)
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## Statement

### Verbal statement

The intersection of a subnormal subgroup with any subgroup is subnormal in that subgroup. Moreover, the subnormal depth of the intersection is bounded from above by the subnormal depth of the original subgroup.

### Statement with symbols

Suppose $H$ is a subnormal subgroup of a group $G$, and $K \le G$ is a subgroup. Then, $H \cap K$ is a subnormal subgroup of $K$. Moreover, if $H$ is $k$-subnormal in $G$, $H \cap K$ is $k$-subnormal in $K$ (i.e., its subnormal depth is at most $k$.

## Facts used

1. Normality satisfies transfer condition: If $H$ is a normal subgroup of a group $G$, and $K \le G$, then $H \cap K$ is a normal subgroup of $K$.
2. Transfer condition is subordination-closed: If $p$ is a subgroup property satisfying the transfer condition, the subordination of $p$.

## Proof

### Hands-on proof

Given: A group $G$, a $k$-subnormal subgroup $H$ of $G$, and a subgroup $K$ of $G$.

To prove: $H \cap K$ is a $k$-subnormal subgroup of $K$.

Proof: Since $H$ is $k$-subnormal subgroup of $G$, we have a subnormal series: $H = H_0 \le H_1 \le \dots \le H_k = G$.

We claim that the following is a subnormal series for $H \cap K$ in $K$: $H \cap K = H_0 \cap K \le H_1 \cap K \le \dots H_k \cap K = K$.

For this, we need to show that each $H_i \cap K$ is normal in $H_{i+1} \cap K$. For this, note that: $H_i \cap K = H_i \cap (H_{i+1} \cap K)$.

Since $H_i$ is normal in $H_{i+1}$, fact (1) tells us that $H_i \cap (H_{i+1} \cap K) = H_i \cap K$ is normal in $H_{i+1} \cap K$, completing the proof.

### Property-theoretic proof

This follows directly from facts (1) and (2).