If $G=G_0 \supsetneq G_1 \supsetneq \cdots \supsetneq \{e\}$ is a composition series for $G$, and if $N \lhd G$, is $G_{i+1}N \lhd G_iN$?

Question

I am stuck understanding some part of a proof and wondering if this is true:

If $G=G_0 \supsetneq G_1 \supsetneq G_2 \supsetneq \cdots \supsetneq \{e\}$ is a composition series for $G$, and if $N \lhd G$, is $G_{i+1}N \lhd G_iN$?

First, I know that $G_{i+1}N$ is a subgroup of $G_iN$ since $N \lhd G_iN$ and $N \lhd G$.

However, can we show that $G_{i+1}N$ is a normal subgroup of $G_iN$?

Answer 1

Let $x \in G_iN$. Then $x=ab$ where $a \in G_i$, and $b \in N$.

Let $y \in G_{i+1}N$. Then $y=gn$ where $g \in G_{i+1}$ and $n \in N$.

Then \begin{align*} xyx^{-1}&=abgnb^{-1}a^{-1}\\ &=abgna^{-1}ab^{-1}a^{-1}\\ &=abgna^{-1}n_1 \quad [\text{since $N \lhd G, \ ab^{-1}a^{-1}\in N $}]\\ &=ag_1b_1na^{-1}n_1 \quad [\text{since $NG_{i+1}=G_{i+1}N, \ bg=g_1b_1$}]\\ &=ag_1a^{-1}ab_1na^{-1}n_1\\ &\in G_{i+1}N. \quad [\text{since $G_{i+1} \lhd G_i, \ ag_1a^{-1}\in G_{i+1}$, and since $ab_1na^{-1}\in N$}] \end{align*}

So, $G_{i+1}N \lhd G_iN$.