The question states:

Let $G$ be an Abelian group with subgroup $H < G$. Let $S=\{x~|~x\in G$ and $x^2 \in H\}$. Show that $S$ is a subgroup of $G$.

My proof is different than what is in the book, I proceeded as follows:

Let the identity be $e$.

$e^2 = e$, therefore $e^2 \in H$, and hence $e \in S$. Therefore $S$ is non-empty.

By definition, $S\subseteq G$.

Let $a$, $b$ $\in S$. Therefore, $a^2, b^2 \in H$ and $a^2, b^2 \in G$.

$\therefore ab^{-1} \in S \iff {(ab^{-1})}^2 \in H$,

$\therefore ab^{-1} \in S \iff a^2(b^2)^{-1} \in H$.

As $H$ is a group, each element has a unique inverse, so $a^2, b^2 \in H \implies b^2 \in H$.

Also, as $H$ is a group, it is closed, and $ \therefore a^2(b^2)^{-1} \in H$. Hence, $ab^{-1} \in S$.

So, S is a non-empty subset of a group G, and $ab^{-1} \in S$ for all $a, b \in S$.

Thus, by the subgroup test, $S$ is a subgroup of $G$.

I never used the fact that G is Abelian though... which is why I am worried my solution may be incorrect.

  • 3
    $\begingroup$ You used the fact that $G$ is abelian in going from $(ab^{-1})^2\in H$ to $a^2(b^2)^{-1}\in H$. You are using $(xy)^2 = x^2y^2$, which is not generally true in nonabelian groups. Yes; your argument seems correct. $\endgroup$ – Arturo Magidin Apr 23 at 4:00
  • $\begingroup$ @ArturoMagidin True, as you can see I'm still having trouble with details such as that which have been ingrained in my brain as absolute truths, and questioning them with group theory, though wonderful, requires me to "rewire" my brain in a sense. So thank you for your observation! $\endgroup$ – John Arg Apr 23 at 4:02
  • $\begingroup$ Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. $\endgroup$ – Shaun Apr 23 at 6:37

To reiterate what was said in the comments by @ArturoMagidin, you used the fact that $G$ is abelian in this step:

$\therefore ab^{-1} \in S \iff {(ab^{-1})}^2 \in H$,

$\therefore ab^{-1} \in S \iff a^2(b^2)^{-1} \in H$.


$$\begin{align} (ab^{-1})^2&=(ab^{-1})(ab^{-1}) &\\ &=a(b^{-1}a)b^{-1}& \\ &=a(ab^{-1})b^{-1} & \text{ (since } G\text{ is abelian),}\\ &=a^2b^{-2}& \\ &=a^2(b^2)^{-1}. & \end{align}$$

Your proof is correct.

The test you use is the "one-step subgroup test", despite there being two steps (because the nonemptiness is ignored for some reason).

Well done!


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