# Equivalent Subnormal Series in a Group

Recall from the Subnormal Series in a Group page that if $G$ is a group and $1$ is the identity in $G$ then a subnormal series in $G$ of length $k$ is a collection of subgroups of $G$ such that:

(1)That is, $G_i$ is a normal subgroup of $G_{i+1}$ for all $i \in \{ 0, 1, ..., k-1 \}$.

We now define what it means for two subnormal series in a group to be equivalent.

Definition: Let $G$ be a group and let $1$ denote the identity in $G$. Two subnormal series $\{ 1 \} = G_0 \trianglelefteq G_1 \trianglelefteq ... \trianglelefteq G_k = G \}$ and $\{ 1 \} = H_0 \trianglelefteq H_1 \trianglelefteq ... \trianglelefteq H_l = G \}$ are said to be Equivalent if:1) $k = l$.2) There exists a bijection $\pi : \{ 0, 1, ..., k-1 \} \to \{ 0, 1, ..., k - 1\}$ such that $G_{i+1} / G_i \cong H_{\pi(i) + 1} / H_{\pi(i)}$ for all $i \in \{ 0, 1, ..., k-1 \}$. |

Therefore, two subnormal series are equivalent if they have the same length and if every factor in the set of factors for the first subnormal series is isomorphic to one of the factors in the set of factors for the second subnormal series.

For example, consider the following subnormal series:

(2)We claim that $(*)$ and $(**)$ are equivalent subnormal series. First observe that $(*)$ and $(**)$ have the same length. We now find the factors of both series. For the subnormal series $(*)$ we have that the factors are:

(4)And for the subnormal series $(**)$ we have that the factors are:

(5)Therefore the subnormal series $(*)$ and $(**)$ are equivalent.