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Subgroups of Z_n when n is even

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
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?
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
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.