# Isomorphism between a quotient group and the 2-torsion subgroup

Let $G$ be a finite abelian group, and let $2G$ denote the subgroup $\{ g * g : g \in G\}$. Let $G$ be the 2-torsion subgroup of $G$.

I want to show that $$G/2G \cong G. \qquad (1)$$

The closest I could get was to prove that $G/G \cong 2G$ using the homomorphism $g \mapsto g*g$ and the First Isomorphism Theorem, but I'm not sure under what criteria it is possible to 'exchange' the two subgroups on either side of the isomorphism sign.

I've tried the most natural homomorphisms I can think of to deduce $(1)$ directly, also using the First Isomorphism Theorem, but none seem to work. Does anyone know the criteria I need (e.g. so that $A/B \cong C \implies A/C \cong B$ under suitable conditions) or could someone perhaps point out a direct homomorphism that does work?

• Certainly one needs something a little special since: B = 4×1 ≤ A = 4×2 and (4×2)/(4×1) ≅ C = 2×1, but (4×2)/(2×1) ≅ 2×2 not 4×1. The criterion I can think of don't apply in your situation, so I think you have to go with Arturo's method. Note that this is not true for infinite 2-groups either: G=Z[1/2]/Z is a 2-group, but G=2G≠G=2. – Jack Schmidt Feb 8 '12 at 22:23

The result is true: first, note that we can write $G$ as the direct sum of its $2$-part and its prime-to-$2$ part: $$G = G_2 \oplus G_{2'},$$ where $$G_2 = \{g\in G\mid g\text{ has order a power of }2\},\qquad G_{2'}=\{g\in G\mid g\text{ has order relatively prime to }2\}.$$ Indeed, if the order of $g$ is $2^mq$ with $q$ odd, then we can express $1 = \alpha 2^m + \beta q$ for some integers $\alpha$ and $\beta$. Thus, $g=(g^q)^{\beta}(g^{2^m})^{\alpha}$, so $g\in\langle g^q\rangle\langle g^{2^m}\rangle$, and $g^q$ has order $2^m$, $g^{2^m}$ has order prime to $2$.

Since $G_{2'}=\{1\}$ and $2G_{2'}=G_{2'}$, the isomorphism on the $2'$ part is trivial: map everything to $1$.

For the $2$-part, write $G_2$ as a direct sum of cyclic groups of order a power of $2$: $$G_2 = C_{2^{a_1}}\oplus\cdots\oplus C_{2^{a_m}},$$ then $2G_2 = 2C_{2^{a_1}}\oplus\cdots\oplus 2C_{2^{a_m}}$, and $G_2=C_{2^{a_1}}\oplus\cdots\oplus C_{2^{a_m}}$. So it suffices to establish the isomorphism in the cyclic of order a power of $2$ case.

And if $C_{2^m}$ is cyclic of order $2^m$, then $C_{2^m}$ is cyclic of order $2$, $2C_{2^m}$ is cyclic of order $2^{m-1}$, and of course we have that $C_{2^m}/2C_{2^m}\cong C_2\cong C_{2^m}$. The map that sends the generator $x$ to $x^{2^{m-1}}$ (the generator of $C_{2^m}$) has kernel $\langle x^2\rangle = 2C_{2^m}$.

Unfortunately, this isomorphism is not canonical/natural once we try to pull it back to $G$ itself, because there are many different ways of writing the decomposition of $G_2$ as a direct sum of cyclic groups of power-of-$2$ order. I think it is best to think of as being in the same "family" as the following theorem:

Theorem. If $G$ is a finite abelian group, and $H$ is a subgroup, then $G$ has a subgroup that is isomorphic to $G/H$.

One can prove this, but there is no "universal" definition of an isomorphism that one can make and that will "always work" to give the isomorphism. It is rather "accidental", as it were.

For example, one can also give the isomorphism by invoking the Structure Theorem: write $G$ as a direct sum of cyclic groups, $$G\cong C_{m_1}\oplus C_{m_2}\oplus\cdots\oplus C_{m_n}$$ where $1\lt m_1|m_2|\cdots|m_n$. Let $x_i$ be the generator of the $i$th cyclic factor. Then define $f\colon G\to G$ by $$f(x_i) = x_i^{m_i/\gcd(m_i,2)}.$$ Again, this maps to $G$, with kernel $2G$: an element $g$ lies in the kernel if and only if its $i$th component is $x_i^a$ and $m_i|m_ia/\gcd(m_i,2)$, if and only if $\gcd(m_i,2)=1$ (in which case $x_i\in 2G$) or $\gcd(m_i,2)=2$ and $2|a$ (in which case $x_i^a\in 2G$), so every component of $g$ lies in $2G$, hence $G$ lies in $2G$ and conversely.

However, note that we cannot make this a general definition; if you attempt to define $$\varphi(x) = x^{m/\gcd(m,2)}, \quad m=|x|$$ then this is not a homomorphism. For example, take $G=C_2\oplus C_4$, generated by $x$ and $y$; then $xy$ and $y$ both have order $4$, so we would have $\varphi(xy) = (xy)^2 = y^2$, $\varphi(y) = y^2$; but $(xy)y=xy^2$ has order $2$, so $\varphi(xy^2)=xy^2$, yet $\varphi(xy)\varphi(y) = y^2y^2 = 1$.

• Thanks for such a quick and in-depth response, Arturo. I have a little addition, though I'm not really expecting a proof of any kind... is it also true that $G = 2G + G$, in the sense of a direct sum of groups? – Sputnik Feb 8 '12 at 23:08
• @Fahad: No; in fact, it is almost never true. For example, in the cyclic-of-order $2^n$ case, $G$ is contained in $2G$ when $n\gt 1$, and of course, no cyclic group of prime power order can be written as a direct sum of two nontrivial groups. Moreover, if $G$ is of exponent $2^n$, then $2G$ is of exponent $2^{n-1}$ and $G$ is of exponent $2$, so $2G\oplus G$ is of exponent $2^{\max(n-1,1)}$, Unless $n=1$ (so $G=G$), you will never have even an isomorphism. – Arturo Magidin Feb 9 '12 at 2:21