2
$\begingroup$

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.
$\endgroup$
0

1 Answer 1

1
$\begingroup$

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 $ah=\underbrace{c}_{\in B}\underbrace{q^{-1}h}_{\in H} \in BH$.)

$\endgroup$
2
  • 1
    $\begingroup$ \bmod, not \mod. \ldots, not ... And \mid gives a vertical line with extra space on both sides. $\endgroup$ Oct 21, 2021 at 16:09
  • $\begingroup$ @ArturoMagidin oh yeah I remembered $\bmod$ but too late to edit. lol. thanks $\endgroup$
    – BCLC
    Oct 21, 2021 at 16:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .