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

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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

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Here is Proposition 10:

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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?

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  • $\begingroup$ To the user casting the close vote, please elaborate on how my question needs clarity. $\endgroup$ – Akira Jul 16 '20 at 15:47
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    $\begingroup$ Hi @Shaun, your appreciation is a great source of encourage for me to study math ^^ $\endgroup$ – Akira Jul 16 '20 at 15:50
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    $\begingroup$ @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.) $\endgroup$ – user1729 Jul 16 '20 at 16:27
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    $\begingroup$ (actually, you should type up the answer yourself!) $\endgroup$ – user1729 Jul 16 '20 at 16:59
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    $\begingroup$ @user1729 I guess the proposition is to ensure "each of these sets of subgroups is closed under intersection". $\endgroup$ – Akira Jul 17 '20 at 9:23
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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 $\phi [S_{1}] \cap \phi [S_{2}] \subseteq \phi[S_1 \cap S_2]$.

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