Show that some group must have a subgroup isomorphic to $(\mathbb{Z}/2\mathbb{Z})^n$ I have two different questions, but they are related.
The first question is, Let $G$ be a finite abelian group. show that if $G$ contains (atleast) $2^n-1$ distinct elements of order 2, then there must be a subgroup of $G$ isomorphic to $(\mathbb{Z}/2\mathbb{Z})^n$.
The second question is, Let $H$ be a finite abelian group.  show that if $H$ contains $\mathfrak{p}_1,...,\mathfrak{p}_n$ which are n distinct elements of order 2, such that for any $I\subsetneq\{1,2,\cdots,n\}$ we have $\prod_{i\in I}\mathfrak{p}_i\neq 1$ Show that $G$ contains a subgroup isomorphic to $(\mathbb{Z}/2\mathbb{Z})^n$.
Some background and context: This may be a fairly easy question to do by induction, but I haven't done group theory in a little bit, and I was never that good at it in the first place. The point of this question is that I was trying to show that a certain quadratic number field had a (ring of integers with a) class number divisible by $2^n$.
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It was pointed out that the second question follows fairly easily from the first one. I believe there should be a relatively easy way to show the first question by induction, but I am struggling with that.
I think the base case is trivial, if a group contains 1 element of order 2, then that element will generate a subgroup isomorphic to $\mathbb{Z}/2\mathbb{Z}$.
So now we assume any (finite abelian) group with (atleast) $2^n-1$ elements of order 2, will have a subgroup isomorphic to $(\mathbb{Z}/2\mathbb{Z})^n$. And let $G$ be a finite abelian group with $2^{n+1}-1$ elements of order 2. We want to show that $(\mathbb{Z}/2\mathbb{Z})^{n+1}$ is isomorphic to a subset of $G$.
I think we can also show pretty easily that, since we have more than $2*(2^n-1)$ elements of order 2, we can find 2 distinct injective homomorphisms $\varphi_1$ and $\varphi_2$ from $(\mathbb{Z}/2\mathbb{Z})^n$ into G, such that their images are disjoint.
So can we just take $\varphi_3$ from $(\mathbb{Z}/2\mathbb{Z})^{n+1}$ to G, and then define it somehow based on the last 2 functions?
 A: For the first question.
Note that in an abelian group $o(xy)|o(x)o(y)$, so that in $G$ the elements whose order is a power of $2$ form a subgroup, and this subgroup has at least $2^n-1$ elements of order $2$. That means we can assume $G$ is a $2$-group.
Now let $\Omega(G)=\{x\in G\mid x^2=1\}$. If $x,y\in g$ we have $(xy)^2=x^2 y^2=1$, so that $\Omega(G)$ is a subgroup and contains all the elements of order $2$. If $\Omega(G)$ has order $2^k$ then it contains $2^k-1$ elements of order $2$, and so $k\geqslant n$.
We can now pick recursively a sequence of non-trivial elements of $\Omega(G)$, $x_1,x_2,\dots, x_k$ such that $x_{j+1}\not\in \langle x_1,x_2,\dots x_j\rangle$ for each $j$: our hypotheses says there are always enough possible choices. Then   $\langle x_1,x_2,\dots x_n\rangle$ is easily seen to be $(\mathbb{Z}_2)^n$.
A: Hint: For your first question: show there is an injective group homomorphism $$\varphi: (\mathbb{Z}/2\mathbb{Z})^n\to G.$$ For the second question: show that the group has $2^n-1$ elements of order $2.$
Added: In order to setup a such group homomorphism, lets look at the  subgroup in $G$ generated by these order two elements. If they were multiplicatively closed then this subgroup is exactly of order $2^n,$ and otherwise bigger than that. In both cases there must be at least $n$ generators for it. So we can simply pick some non-trivial distinct $n$ generators and look at (possibly smaller) subgroup generated by them. If you write additively, say, $$H=\langle e_i: 2e_i=0, 1\le i\le n\rangle.$$ A general element $h\in H$ is of the form $$h=\sum_{1\le i \le n}\epsilon_ie_i$$ where $\epsilon_i$ is either zero or one, hence $|H|=2^n$. Now we can simply define the desire group homomorphism by
$$\varphi((\underbrace{0, 0,\cdots,1,\cdots, 0}_{i\text{-th coordinate is one}}))=e_i$$ and extending linearly. Clearly this is a homomorphism and by construction its image is $H.$
Hence a bijection onto the image.
