# A doubt about Theorem 22 in textbook Algebra by Saunders MacLane and Garrett Birkhoff

I'm reading Theorem 22 in textbook Algebra by Saunders MacLane and Garrett Birkhoff.

It follows that $$\phi_*S = \phi[S] := \{\phi(x) \mid x \in S\}$$ and $$\phi^*T = \phi^{-1}[T] := \{x \in G \mid \phi(x) \in T\}$$.

and its proof

Here is Proposition 10:

Due to the properties of the set-valued functions $$\phi[\cdot], \phi^{-1}[\cdot]$$ induced from $$\phi (\cdot)$$, we always have $$\phi_{*}\left(S_{1} \cap S_{2}\right) \subseteq \phi_{*} S_{1} \cap \phi_{*} S_{2}$$. One sufficient condition for the equality to hold is that $$\phi$$ is injective.

Could you please elaborate on how Proposition 10 lead to the equality?

• To the user casting the close vote, please elaborate on how my question needs clarity. – Akira Jul 16 '20 at 15:47
• Hi @Shaun, your appreciation is a great source of encourage for me to study math ^^ – Akira Jul 16 '20 at 15:50
• @LAD Oh, right. For this direction you need to understand and use the kernel. In particular, for arbitrary subgroups $S\leq G$ we have $\phi^{-1}(\phi(S))=S\ker\phi$, and here the kernel is contained in what you will use for $S$, and therefore that S\ker(\phi)=S\$. (Sorry, don't have time to flesh out the details.) – user1729 Jul 16 '20 at 16:27
• (actually, you should type up the answer yourself!) – user1729 Jul 16 '20 at 16:59
• @user1729 I guess the proposition is to ensure "each of these sets of subgroups is closed under intersection". – Akira Jul 17 '20 at 9:23

Clearly, $$\phi [S_{1} \cap S_{2}] \subseteq \phi [S_{1}] \cap \phi [S_{2}]$$. Below is my use of kernel to obtain $$\phi [S_{1}] \cap \phi [S_{2}] \subseteq \phi [S_{1} \cap S_{2}]$$:

For $$y \in \phi[S_1] \cap \phi[S_2]$$, $$y=\phi(x_1)=\phi(x_2)$$ for some $$x_1 \in S_1, x_2 \in S_2$$. Then $$\phi(x_1 x_2^{-1}) = \phi(x_1) \phi(x_2)^{-1} =1$$. Hence $$x_1 x_2^{-1} \in \operatorname{ker} \phi \subseteq S_1 \cap S_2$$ and thus $$x_1 x_2^{-1} \in S_1$$. Because $$S_1$$ is a subgroup, $$x_2 \in S_1$$. Hence $$x_2 \in S_1 \cap S_2$$. The result then follows.

Here is a lemma suggested by @user1729.

Lemma: If $$\phi:G \to H$$ is morphism of groups and $$S$$ is a subgroup of $$G$$, then $$\phi^{-1}[\phi[S]] = S \operatorname{ker} \phi = S$$.

Proof: Notice that $$\phi^{-1}[\phi[S]] = \{x \in G \mid \exists y\in S: \phi(x) = \phi (y)\} \overset{(\star)}{=} \{x \in G \mid \exists y\in S: xy^{-1} \in \operatorname{ker} \phi\}$$. It follows that $$\phi^{-1}[\phi[S]] = S \operatorname{ker} \phi$$. Notice that $$S \subseteq\phi^{-1}[\phi[S]]$$. With similar reasoning in my above approach, we get $$(x,y) \in S \times G$$ and $$\phi(x) = \phi(y)$$ implies $$y \in S$$. Hence $$\phi^{-1}[\phi[S]] = S$$.

$$(\star)$$: This is because $$\phi$$ is a morphism of groups.

Then we use this lemma to obtain the latter inclusion as follows:

We have $$\phi [S_{1}] \cap \phi [S_{2}] \subseteq \phi [S_{1}]$$ and thus $$\phi^{-1}[\phi [S_{1}] \cap \phi [S_{2}]] \subseteq \phi^{-1}[\phi [S_{1}]] \color{red}{=} S_1$$. Similarly, $$\phi^{-1}[\phi [S_{1}] \cap \phi [S_{2}]] \subseteq S_2$$. Hence $$\phi^{-1}[\phi [S_{1}] \cap \phi [S_{2}]] \subseteq S_1 \cap S_2$$ and thus $$\phi [S_{1}] \cap \phi [S_{2}] \subseteq \phi[S_1 \cap S_2]$$.