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In the following problems, let $G$ be an Abelian group.

1) Let $H = \{ x \in G: x=y^{2} \text{ for some } y \in G \}$; that is, let $H$ be the set of all the elements of $G$ which have a square root. Prove that $H$ is a subgroup of $G$.

(i). Let $a, b \in H$, then $a = c^{2}$ and $b = d^{2}$ for some $c$ and $d \in G$. The product $ab = c^{2}d^{2}$ shows that $ab$ has a square root because $c^{2}d^{2} = ccdd =(cd)(cd) = (cd)^{2}$. Because $G$ is a group, it is closed under multiplication so $cd \in G$.

(ii). Let $a \in H$, then $a = c^{2}$ for some $c \in G$. Since $c \in G, c^{-1} \in G$ because $G$ is a group.

This is where I am confused:

Am I allowed to make this conclusion: $a^{-1} = (c^{2})^{-1} = (c^{-1})^{2}$ which shows that $a^{-1} \in H$?


Comments on my next proof would be appreciated.

2) Let $H$ be a subgroup of $G$ and let $K = \{ x \in G: x^{2} \in H \}$. Prove that $K$ is a subgroup of $G$.

(i). Let $a, b \in K$, then $a, b \in G$ and $a^{2}, b^{2} \in H$. We need to show that $ab \in G$ and $(ab)^{2} \in H$. Since $G$ is a group, it must be closed with respect to group multiplication. Since $a, b \in G$, the product $ab \in G$ as well. Since $H$ is a subgroup, it must also be closed with respect to multiplication. Since $a^{2}, b^{2} \in H$, the product $a^{2}b^{2} = aabb = (ab)^{2} \in H$ making use of the fact that $G$ is Abelian.

(ii.) In order to show that $K$ is a subgroup of $G$, it is necessary to show that for every element $a$, there is an inverse element $a^{-1}$ such that $a^{-1} \in G$ and $(a^{-1})^{2} \in H$. Because $G$ is a group, it is closed with respect to inverses. So for every element $a$ there is an inverse element $a^{-1} \in G$. Also since $a^{2} \in H$ and $H$ is a subgroup, it is closed to inverses. So $(a^{2})^{-1} = (a^{-1})^{2} \in H$.

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    For your first question, you are indeed allowed to draw that conclusion: if $a=c^2$, then $a(c^{-1})^2=c^2(c^{-1})^2=1_G$, so $a^{-1}=(c^{-1})^2\in H$.2011-08-10
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    @Brian: That clears it up. Thank you.2011-08-10

1 Answers 1

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(1) Don't just use that $(c^2)^{-1} = (c^{-1})^2$, prove it! Note that $(c^2)^{-1}=(c^{-1})^2$ if and only if $(c^{-1})^2$ is the unique element $x$ of $G$ such that $x(c^2) = 1$. So show that $(c^{-1})^2$ has this property to establish it.

Also: don't forget to note that $H$ is nonempty (this is easy, but nonetheless an important part of showing a subset is a subgroup). This is also missing in (2): you need to show that $K$ is not empty.

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    If I want to show that $H$ is not empty, is it enough to let $H = \{4, 9, 16 \}$ and let $G$ be the set of all real numbers? In general, when I am looking for examples to show that $H$ is not empty, do I need to choose examples such that the product of any two elements in $H$ will be in $G$ and that the inverse of any element in $H$ will be in $G$?2011-08-10
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    @Jon: What you propose doesn't work for several reason: (i) If $G$ is the set of all real numbers, then presumably you are looking at the *additive group*, and then $H = G$, so $H=\{4,9,16\}$ is incorrect; even if you are looking at the multiplicative group, $H$ is **still** not equal to what you propose. And (ii) Even if this were correct, you would only have shown that $H$ is not empty for *this particular $G$*. But the problem is for an *arbitrary* $G$, so this certainly does not work. To show that $H$ is not empty, just exhibit an element that is in $H$! e.g., $e\in H$ because $e=e^2$.2011-08-11
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    @Jon: Note that if you are trying to prove that something is a subgroup, it better contain the identity. So **most of the time** the simplest way to show that a given subset-that-you-are-trying-to-prove-is-a-subgroup is not empty is to note/observe that the identity of the group lies in the set. *Sometimes* this is not entirely obvious (seldom the case), in which case you just need to exhibit *some* element that necessarily lies in $H$.2011-08-11
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    @Jon (cont) Here, to show $H$ is not empty (no matter what $G$ is) just observe that since $e^2 = e$, and $e\in G$, then $e\in H$; to show that $K$ is not empty, you can likewise note that since $e^2\in H$, then it follows that $e\in K$.2011-08-11
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    Your explanation clears it up. Thank you.2011-08-12