# Proof of Goursat's lemma in Serre's Finite Groups An Introduction

On page 8 in Serre's Finite Groups An Introduction about the proof of Goursat's lemma it states that:

Proposition 1.6 (Goursat's lemma). For $$i=1,2$$ , let $$N_i$$ be a normal subgroup of $$G_i$$ and let $$\varphi : G_1/N_1\to G_2/N_2$$ be an isomorphism. Let $$H_{N_1,N_2,\varphi}$$ be the set of all $$(g_1,g_2)\in G_1\times G_2$$ which are such that $$\varphi(\bar{g_1})=\bar{g_2}$$ where $$\bar{g_i}$$ is the image of $$g_i$$ in $$G_i/N_i$$ . Then:

i) $$\text{pr}_i(H_{N_1,N_2,\varphi})=G_i$$ for $$i= 1,2$$ .

ii) Every subgroup $$H$$ of $$G_1\times G_2$$ such that $$\mathrm{pr}_i(H)= G_i$$ for $$i = 1,2$$ is equal to some $$H_{N_1,N_2,\varphi}$$ for a unique choice of $$N_1,N_2$$ and $$\varphi$$ .

[Note that $$H_{N_1,N_2,\varphi}$$ may be viewed as a kind of graph of $$\varphi$$ .]

Proof of ii). Define $$N_1$$ as $$H\cap G_1$$ , where $$G_1$$ is viewed as a subgroup of $$G_1\times G_2$$ , namely the kernel of $$\text{pr}_2$$ . Define similarly $$N_2 = H\cap G_2$$ . By assumption, we have $$H.G_1=G_1\times G_2$$ , since both $$H$$ and $$G_2$$ normalize $$H$$ and $$G_1$$ , this shows that $$G_1\times G_2$$ normalizes $$N_1$$ , i.e., that $$N_1$$ is normal in $$G_1$$ . Similarly $$N_2$$ is normal in $$G_2$$ . $$\cdots$$

In the proof of part ii , the author says that "since both $$H$$ and $$G_2$$ normalize $$H$$ and $$G_1$$ ", but I can not see how to show that $$G_2\leqslant\mathbf{N}_{G_1\times G_2}(H)$$ . If so, then similarly we have $$G_1\leqslant\mathbf{N}_{G_1\times G_2}(H)$$ . Pick arbitrary $$(g_1,1)\in G_1$$ and $$(1,g_2)\in G_2$$ we obtain $$(g_1,g_2)=(g_1,1)(g_2,1)\in\mathbf{N}_{G_1\times G_2}(H)$$ since both $$G_1$$ and $$G_2$$ normalize $$H$$ , this forces $$G_1\times G_2=\mathbf{N}_{G_1\times G_2}(H)$$ , then $$H$$ must be normal in $$G_1\times G_2$$ .

• I would replace "$H$ and $G_1$" by "$H\cap G_1$". Commented Mar 15 at 2:32

I think this is perhaps less than clear and you are misinterpreting what is meant. Rather than "$$H$$ and $$G_1$$", it's probably best to say that $$H$$ normalizes both $$H$$ and $$G_1$$, and $$G_2$$ centralizes $$G_1$$, hence normalizes $$H\cap G_1$$. I also think there is a typo there.
Explicitly: we want to show $$H\cap(G_1\times\{1\})$$ is normal. Note $$G_2$$ centralizes $$G_1\times\{1\}$$, so it centralizes (hence normalizes) any subgroup of $$G_1\times\{1\}$$; in particular it normalizes $$H\cap (G_1\times\{1\})$$. And $$H$$ normalizes both $$H$$ and $$G_1\times\{1\}$$ (the latter because it is normal in $$G_1\times G_2$$). So $$H$$ normalizes $$H\cap (G_1\times\{1\})$$. Thus, both $$H$$ and $$G_2$$ normalize $$H\cap (G_1\times \{1\})$$.
Now, (typo alert) we have $$HG_2=G_1\times G_2$$. Since $$H$$ and $$G_2$$ both normalize $$H\cap(G_1\times\{1\})$$, and $$H$$ and $$G_2$$ generate $$G_1\times G_2$$, it follows that $$H\cap (G_1\times\{1\})$$ is normal in $$G_1\times G_2$$. The symmetric argument, using $$G_1$$ instead of $$G_2$$ and that $$HG_1=G_1\times G_2$$, shows that $$H\cap(\{1\}\times G_2)$$ is normal in $$G_1\times G_2$$.
• Do you you mean that $H$ does not necessarily normalize $G_1$? I have finished the proof following your leading. Thanks! Commented Mar 15 at 0:53
• @DianWei $H$ normalizes both $G_1$ and $G_2$, because both $G_1$ and $G_2$ are normal in $G_1\times G_2$, and $H\leq G_1\times G_2$. I think you mean to ask whether $G_1$ (or $G_2$) normalize $H$. They need not. Take $G_1=G_2=S_3$, and take $H$ to be the diagonal subgroup $H=\{(x,x)\mid x\S_3\}$. Then $H$ is not normalized by either $S_3\times\{1\}$ nor by $\{1\}\times S_3$. Commented Mar 15 at 0:56
• Yes, I had a doubt on whether $G_2\leqslant\mathbf{N}_{G_1\times G_2}(H)$ , it is a typo in my initial comment. Thanks for your counter-example. Commented Mar 15 at 11:06