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I'm trying to show the last part to the following question:

Every element in a group $G$ has order $2$, prove $G$ is abelian. Show that if $H$ is a subgroup of G and $g \in G\backslash H$ then $K = H \cup gH$ is a subgroup of $G$. Show further that $K$ isomorphic to $H \times C_2$. Deduce that if $G$ is finite then $G$ is isomorphic to $(\mathbb{Z}_2)^n$ for some non-negative integer $n$

Ok, so I've done all but the very last part where you show $G$ is isomorphic to $(\mathbb{Z}_2)^n$ if it is finite. Just in case there's any notational ambiguity, $C_2$ is the second cyclic group and $(\mathbb{Z}_2)^n$ is just another way to write $(\mathbb{Z}/2\mathbb{Z})^n$.

My line of thinking is that since $G$ is abelian, we know that the left- and right-cosets are equal, I used this fact in showing that $K \cong H \times C_2$. Now, because the left and right cosets also partition a group, am I right in saying that since $gH = Hg$, this implies that $K = H \cup gH = H \cup Hg = H \cup G \backslash H = G$? So, since $K \cong H \times C_2$ this implies that $G \cong H \times C_2$. Since $H$ is a subgroup of $G$, all elements in $H$ are isomorphic to $\mathbb{Z}_2$ as they have order $2$ and clearly $C_2 \cong \mathbb{Z_2}$, hence, as $G$ is finite, for some integer $p$

$G \cong (\mathbb{Z}_2)^p \times \mathbb{Z}_2 = (\mathbb{Z}_2)^{p+1}$

Upon letting $n = p +1$ we get that

$G \cong (\mathbb{Z}_2)^n$

Is (any) of this correct? As I feel I might be barking up the wrong tree. Thanks for any help!

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1 Answer 1

up vote 3 down vote accepted

If I had to follow the hints, I would argue by induction. I would start with a subgroup $$H_1 = \langle a_1 \rangle \cong C_2,$$ where $1 \ne a_1 \in G$, then take $a_2 \in G \setminus H_1$, so that $$H_2 = H_1 \cup a_2 H_1 \cong H_1 \times C_2 \cong C_2 \times C_2,$$ then take $a_3 \in G \setminus H_2$, etc.

However, if I were free to choose my proof, I would prefer to show that $G$ can be regarded as a $\Bbb{Z}_2$-module, that is, a finite-dimensional vector space over $\Bbb{Z}_2$, so that $G$ is isomorphic to the additive group of the vector space $\Bbb{Z}_{2}^{n}$, where $n$ is the dimension.

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Ah yes, this is very clever. Am I incorrect in my assumption that if $G$ is abelian, then $gH = Hg$ implies that $gH = Hg = G \backslash H$? –  Noble. Mar 18 '13 at 12:21
    
@Noble., sorry, I missed to comment on that. Suppose your group has order at least $8$. Start with $H$ of order $2$, and let $g \in G \setminus H$. Then $H \cup g H$ is a subgroup of $G$ of order $4$, so $H \cup g H \subsetneq G$ and $g H \subsetneq G \setminus H$. –  Andreas Caranti Mar 18 '13 at 12:24
    
Ok thanks, so I can't use that in my proof. It definitely seems like induction is the way to go, in light of this. –  Noble. Mar 18 '13 at 12:27
    
Sorry, there's one last thing that's just confusing me now. How do we conclude, by induction, that $G$ is isomorphic to $(\mathbb{Z}_2)^n$ from keep taking an element from $G$ and using $K$. Is it because $K$ is a subgroup, so if we keep expanding the subgroup with another element in $G$ we'll eventually have $G$ itself? Thanks again. –  Noble. Mar 18 '13 at 12:55
    
@Noble., yes, I am taking an increasing sequence of subgroups $1 < H_1 < H_2 < \dots$. Since $G$ is finite, we will eventually reach $G$. –  Andreas Caranti Mar 18 '13 at 12:59

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