Is it true that $x (\bigcap\limits_{\alpha \in A} T_\alpha) \neq \bigcap_{\alpha \in A} (x T_\alpha)$?

Question

I want to show that the intersection of normal subgroups are normal subgroup in direct way.

It means that if $T_\alpha$'s are normal subgroups of a group $G$, then I want to show that

$$x (\bigcap\limits_{\alpha \in A} T_\alpha) x^{-1} = \bigcap_{\alpha \in A} (x T_\alpha x^{-1}) ~~ (= \bigcap_{\alpha \in A} T_\alpha)$$ for every $x \in G$.

To do this, first I need the identity
$$x (\bigcap\limits_{\alpha \in A} T_\alpha) = \bigcap_{\alpha \in A} (x T_\alpha).$$

We can easily show that $$x (\bigcap\limits_{\alpha \in A} T_\alpha) \subset \bigcap_{\alpha \in A} (x T_\alpha).$$

However, I'm stuck to show the opposite inclusion:

$$\bigcap_{\alpha \in A} (x T_\alpha) \subset x (\bigcap\limits_{\alpha \in A} T_\alpha).$$

Is this statement true? Or is there a counterexample?

Answer 1

Consider the function $\lambda_x \colon G \to G$ defined by $\lambda_x(y) = xy$. Note that if $S$ is any subset of $G$, then its direct image under $\lambda_x$ is $xS$, that is, $$\lambda_x[S] = xS.$$

Since the direct image commutes with intersections when the function is injective, we have $$ x \left( \bigcap_{\alpha \in A} T_\alpha \right) = \lambda_x \left[ \bigcap_{\alpha \in A} T_\alpha \right] = \bigcap_{\alpha \in A} \lambda_x[T_\alpha] = \bigcap_{\alpha \in A} (xT_\alpha). $$ In general, viewing these kind of equalities as the typical direct/inverse image identities can be useful for deciding whether they are true or not.

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Leaving this aside, a direct proof of $\bigcap_{\alpha \in A} (xT_\alpha) \subseteq x \left( \bigcap_{\alpha \in A} T_\alpha \right)$ is easy:

Take $y \in \bigcap_{\alpha \in A} (xT_\alpha)$ and $\alpha \in A$. As $y \in xT_\alpha$, we have $x^{-1}y \in T_\alpha$. Since $\alpha$ was arbitrary, $x^{-1}y \in \bigcap_{\alpha \in A} T_\alpha$, and then $y \in x \left( \bigcap_{\alpha \in A} T_\alpha \right)$.