Subgroups of Z_n when n is even

[alyafey22]

Contemporary Abstract Algebra by Gallian

This is Exercise 24 Chapter 3 Page 70

Question

Suppose $n$ is an even positve integer and $H$ is a subgroup of $\mathbb{Z}_n$. Prove that either every member of $H$ is even or exactly half of the members of $H$ are even
.

Attempt

We consider some cases :

[1] The subgroup generated by the identity $0$ then $H=<0>=\{0\}$ hence all members are even .
[2] The subgroup generated by $1$ , $H=G$ and since the ordere of $G$ is even , we have half the members are even.
[3] The subgroup generated by $2$, $H=<2>=\{0,2,4,6 , \cdots , 2n\}$
hence all members are even.
[4] The subgroup generated by $n$ , $H=<n>=\{0,n\}$. If $n$ is even then subgroup contains even integers. If $n$ is odd half the members are even.

Since $\mathbb{Z}_n$ is cyclic every subgroup is cyclic. So it suffices to look at the subgroups generated by each even and odd element. I am taking this theorem from another latter chapter so I guess I shouldn't be able to use the properties of cyclic groups.

Any hint how to proceed ?

 

[Deveno]

The theorem stated is equivalent to this one:

Let $K$ be the subgroup of "even" elements. Then either:

1. $|K \cap H| = |H|$

or:

2. $|K \cap H| = \dfrac{|H|}{2}$.

Some observations about your observations:

Your description of the subgroup generated by $n$ is incorrect: in $\Bbb Z_n$ we have that:

$[n] = [0]$, since $n \equiv 0 \text{ (mod } n)$.

Before I go any further, might I ask: have you covered Lagrange's theorem yet?

 

[alyafey22]

Some observations about your observations:

Your description of the subgroup generated by $n$ is incorrect: in $\Bbb Z_n$ we have that:

$[n] = [0]$, since $n \equiv 0 \text{ (mod } n)$.

I actually abused the notations. Since $n$ is even we have $n=2k$ for some positive $k$ then we have the subgroup generated by $k$ contains $\{0,k\}$.

Before I go any further, might I ask: have you covered Lagrange's theorem yet?

No, we only covered groups ,subgroups tests and examples of subgroups.

 

[Deveno]

Re: Supgroups of Z_n when n is even

OK. In this case, note that "even" only makes sense when $n$ is an even integer, for if $n$ is odd, "skipping by 2's" lands on every element.

Also note that we can't use our "usual" definition of "even": that $k$ is even if 2 divides $k$, because "divides" doesn't always make sense in $\Bbb Z_n$ (consider $n = 6$ and what "division by 3" could possibly mean).

SO we have to use THIS definition:

$[k]$ is even if $[k]$ is in the subgroup generated by $[2]$ which is:

$\{[0],[2],[4],...,[n-2]\}$.

Equivalently, $[k]$ is even if $k$ is even.

Now, let's do this:

Suppose $H = \{[0],[k_1],...,[k_r]\}$ with:

$1 \leq k_1 <\dots < k_r < n$.

Claim 1:

$H$ consists solely of sums:

$[k_1], 2[k_1],\dots,s[k_1] = [0]$ (here, $t[k_1]$ is a $t$-fold sum of $k_1$ with itself).

Clearly any such sum is in $H$ by closure. Since we have an infinite number of such sums, and $H$ is finite, it must be that for two positive integers $t_1 < t_2$:

$t_1[k_1] = t_2[k_1]$, so $(t_2 - t_1)[k_1] = [0]$.

Since any non-empty set of positive integers has a least element, we have a least such positive integer, $s$. I claim that $s$ divides $n$ and moreover, that $k_1$ does, as well.

Write $n = qk_1 + r$, where $r = 0$ or $r < k_1$.

In $H$, this becomes:

$[0] = q[k_1] + [r]$, showing that $[r] \in H$. Since we chose $k_1$ to be minimal, $r = 0$.

This shows that $k_1$ divides $n$.

Now suppose $k_j$ is not a multiple of $k_1$. Again, we can write:

$k_j = q'k_1 + r'$, and by the same reasoning, we must have $r' = 0$, which contradicts our assumption that $k_j$ is not a multiple of $k_1$.

But now we see that the smallest positive integer $s$ for which $s[k_1] = [0]$ (in $\Bbb Z_n$) must be the $q$ we posited earlier, since for every smaller positive integer $t$, we have:

$tk_1 = k_j < n$.

This proves claim 1.

We now consider two cases:

Case 1: $k_1$ is even.

In this case, we see every element of $H$ is even.

Case 2: $k_1$ is odd. In this case, since $sk_1 = n$ and $n$ is even, We must have an even number of elements in $H$, namely:

$[k_1],2[k_1],\dots,(s-1)[k_1],[0]$.

Clearly, every other element is even, and there are $s/2$ such elements.

 

[blue_raver22]

One approach to proving this statement is to use the fact that every subgroup of $\mathbb{Z}_n$ is of the form $\langle d \rangle$ for some divisor $d$ of $n$. Since $n$ is even, we know that $d$ must also be even. Therefore, every subgroup of $\mathbb{Z}_n$ is either generated by an even element (in which case all members are even), or by an odd element (in which case exactly half of the members are even). This shows that the statement is true for all subgroups of $\mathbb{Z}_n$ when $n$ is even.