# finitely generated quotient group: From $G/H = \langle g_1H, \ldots, g_m H \rangle$, we have $G=\langle g_1, \ldots, g_m \rangle H$?

Please help me with my own proof for this: A group with finitely generated normal subgroup and finitely generated quotient is finitely generated itself

Let $$G$$ be a group with $$G \trianglerighteq H$$ normal subgroup. Assume that $$H$$ is finitely generated, and $$G / H$$ (the quotient group) is finitely generated as well. Is $$G$$ finitely generated? (Answer: Yes.)

From finite generation we have $$H=\langle h_1, ..., h_n \rangle$$ and $$G/H = \langle g_1H,\ldots, g_m H \rangle$$ for some $$n,m \ge 1$$. I've come up with my own proof based on the hints in previous question. I would like to please clarify this part:

From $$G/H = \langle g_1H, \ldots, g_m H \rangle$$, we have $$G=\langle g_1, \ldots, g_m \rangle H$$.

Questions:

1. Is the following what is going on?

We are given: $$G/H = \langle g_1 \bmod H, \ldots, g_m \bmod H \rangle$$

Then we deduce: $$G=\langle g_1,\ldots, g_m \rangle H$$. This is now product set and not $$\bmod$$ or anything.

1. So how do I prove (1) exactly? Here's what I tried:

Let $$\pi_H$$ be the canonical map ($$\pi_H: G \to G/H, \pi_H(g)= gH$$ as in $$g \bmod H$$).

$$\pi_H(G) = G/H = \langle g_1 \bmod H, \ldots, g_m \bmod H \rangle$$

$$\{(g_1 \bmod H)^{k_1} (g_2 \bmod H)^{k_2} \cdots (g_m \bmod H)^{k_m}\mid k_1, \ldots, k_m \in \mathbb Z\}$$

$$= \{(g_1^{k_1} \bmod H) (g_2^{k_2} \bmod H) \cdots (g_m^{k_m} \bmod H)|k_1, \ldots, k_m \in \mathbb Z\}$$

$$= \{g_1^{k_1}g_2^{k_2} \cdots g_m^{k_m} \bmod H\mid k_1, ldots, k_m \in \mathbb Z\}$$

$$= \{\pi_H(g_1^{k_1}g_2^{k_2} \cdots g_m^{k_m})\mid k_1, \ldots, k_m \in \mathbb Z\}$$

$$= \pi_H(\{(g_1^{k_1}g_2^{k_2} \cdots g_m^{k_m}\mid k_1, \ldots, k_m \in \mathbb Z\})$$

$$= \pi_H(\langle g_1,\ldots, g_m \rangle)$$

And then

$$\pi_H(\langle g_1, \ldots, g_m \rangle) = \{g \bmod H \mid g \in \langle g_1, \ldots, g_m \rangle \}$$

while

$$\pi_H(G) = G/H = \{g \bmod H | g \in G \}$$

Not sure how to conclude from here. Is there some rule like for elements $$a,b \in G$$, we get from $$\pi_H(a) = \pi_H(b)$$ that $$a \bmod H = b \bmod H$$ but for subsets/subgroups $$A,B \subseteq G$$, we get from $$\pi_H(A) = \pi_H(B)$$ that $$AH=BH$$? (well perhaps on the element level we get $$\{a\}H=\{b\}H$$)

• Edit 1: Got it I think. See an answer I posted. Actually I made mistakes above
1. in saying that

$$\langle g_1 \bmod H, \ldots, g_m \mod H \rangle = \{(g_1 \bmod H)^{k_1} (g_2 \bmod H)^{k_2} \cdots (g_m \bmod H)^{k_m}\mid k_1, \ldots, k_m \in \mathbb Z\}$$

1. in assuming that $$\{(g_1^{k_1}g_2^{k_2} \cdots g_m^{k_m}\mid k_1, \ldots, k_m \in \mathbb Z\}$$ (when I was saying that $$\pi_H(\{(g_1^{k_1}g_2^{k_2}\cdots g_m^{k_m}\mid k_1, \ldots, k_m \in \mathbb Z\}) = \pi_H(\langle g_1, \ldots, g_m \rangle)$$).

1. Is it nonsensical to continue the argument with something like the following?

From $$G/H = \langle g_1H,\ldots, g_m H \rangle$$, we have $$G=\langle g_1, \ldots, g_m \rangle H$$.

Then

$$G=\langle g_1, \ldots, g_m \rangle H$$

$$=\langle g_1, \ldots, g_m \rangle \langle h_1, \ldots, h_n \rangle$$

$$=$$ something like $$\langle g_1, \ldots, g_m, h_1, \ldots, h_n \rangle$$ or $$\langle \{g_ih_j \mid i=1,\ldots,m; j=1,\ldots, n\} \rangle$$

• Edit 2: I think it's sensible because the product set $$\langle g_1, \ldots, g_m \rangle \langle h_1, \ldots, h_n \rangle$$ is given to be equal to a subgroup of $$G$$ (namely the whole of $$G$$) and thus we can just combine the indices/generators.

Wait after overcoming my fear of using $$gH$$ when what is meant is $$g \mod H$$, I think I figured it out. Might as well just type it here as an answer.

To prove $$G = \langle g_1, \ldots, g_m\rangle H$$:

$$\supseteq$$ duh

$$\subseteq$$ Let $$g \in G.$$ Then $$gH = \pi_H(g) \in \pi_H(G)=\pi_H(\langle g_1, \ldots, g_m \rangle) = \{kH\mid k \in \langle g_1, \ldots, g_m \rangle \}$$. Then $$gH=kH$$, for some $$k \in \langle g_1, \ldots, g_m \rangle$$. Hence, $$gh=kl$$, for some $$h,l \in H$$. Therefore, $$g = \underbrace{k}_{\in \langle g_1, \ldots, g_m \rangle}\underbrace{lh^{-1}}_{\in H}$$

Wait I also figured out $$AH=BH$$ (given $$\pi_H(A) = \{a \bmod H\mid a \in A\} = \{b \bmod H\mid b \in B\} = \pi_H(B)$$) even if you use $$\bmod H$$:

$$\supseteq$$ By symmetry, 1 direction is sufficient.

$$\subseteq$$ Let $$ah \in AH$$, for $$(a,h) \in A \times H$$. We must show that $$ah=bk$$, for some (b,k) \in B \times H.

Now, when $$a \bmod H = \pi_H(a) \in \pi_H(A) = \pi_H(B) = \{b \bmod H\mid b \in B\}$$. Then $$a \bmod H = c \bmod H$$, for some $$c \in B$$. Hence, $$ca^{-1}=q$$, for some $$q \in H$$. Therefore, $$ah=cq^{-1}h$$. Choose $$(b,k)=(c,q^{-1}h)$$. (or simply note that $$ah=\underbrace{c}_{\in B}\underbrace{q^{-1}h}_{\in H} \in BH$$.)

• \bmod, not \mod. \ldots, not ... And \mid gives a vertical line with extra space on both sides. Oct 21, 2021 at 16:09
• @ArturoMagidin oh yeah I remembered $\bmod$ but too late to edit. lol. thanks
– BCLC
Oct 21, 2021 at 16:11