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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 $a^{-1}=(c^{-1})^2\in H$. –  Brian M. Scott Aug 10 '11 at 20:22
    
@Brian: That clears it up. Thank you. –  Student Aug 10 '11 at 20:33
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1 Answer

up vote 3 down vote accepted

(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 $G$? –  Student Aug 10 '11 at 21:05
    
@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 $e=e^2$. –  Arturo Magidin Aug 11 '11 at 3:56
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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 $H$. –  Arturo Magidin Aug 11 '11 at 4:17
    
@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 $e\in K$. –  Arturo Magidin Aug 11 '11 at 4:18
    
Your explanation clears it up. Thank you. –  Student Aug 12 '11 at 20:41
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