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A Subgroup of Integers

In A Ring of Mappings we were able to show that the set of mappings on \(\mathbb{R}\) formed a group under addition. Here, we will look more into groups as we show that the multiples of an integer \(n\) forms a subroup of the group of Integers under addition.

A Subgroup

Show that \(n\mathbb{Z}\) is a subroup of \(\mathbb{Z}\) under addition for all \(n \in \mathbb{Z}\).

Definitions

Before we get into the proof, we want to make sure we understand what it is we are trying to show. To begin with, we need to recall that
\begin{align*}
n\mathbb{Z}=\left\{nk: k \in \mathbb{Z}\right\}.
\end{align*}
That is, \(n \mathbb{Z}\) is the set of all integer multiples of \(n\). Now, if \(n\) is an integer, we should notice that all integer multiples of \(n\) will be integers since the integers are closed under multiplication. However, if \(n\) is not an integer, this is not the case. Therefore, we will have that \(n \mathbb{Z}\) is a subset of \(\mathbb{Z}\) if and only if \(n \) is an integer.

Next, we need to recall that a subgroup is a subset of a given group which is again a group. Since we have seen that \(n\mathbb{Z}\) is a subset of \(\mathbb{Z}\) as long as \(n\) is an integer, so we now just need to show that \(n\mathbb{Z}\) is itself group.

Now, we could go through step by step to show this as we did in A Ring of Mappings. However we can use the following lemma,

Lemma:
A non-empty subset, \(G\), of a group is subgroup if and only if for all \(a,b \in G\) we have that \(ab^{-1} \in G\).

That is, in this situation, we have to show that, for every \(a,b \in n\mathbb{Z}\) it holds that \(a-b \in n \mathbb{Z}\). If we take two elements in \(n \mathbb{Z}\), say \(a,b\), we then look at \(a-b\).

We first note that \(n\mathbb{Z} \neq \emptyset\) since \(0=n*0 \in n\mathbb{Z}\). Furthermore, if \(a,b \in n \mathbb{Z}\), we have that \(a=nk_{1}\) and \(b=nk_{2}\) for some \(k_{1},k_{2} \in \mathbb{Z}\). Hence,
\begin{align*}
a-b &=nk_{1}-nk_{2} \\
&=n(k_{1}-k_{2}).
\end{align*}
Since \(k_{1}-k_{2} \in \mathbb{Z}\), this tell us that \(a-b \in n\mathbb{Z}\). Hence, \(n \mathbb{Z}\) is a subgroup of \(\mathbb{Z}\) for addition.

We can now put this together for a formal proof.

Proof

Let \(n \in \mathbb{Z}\). We note that \(0=n*0 \in n\mathbb{Z}\), so \(n \mathbb{Z} \neq \emptyset\). Now, let \(a,b \in n \mathbb{Z}\). Then we have that \(a=nk_{1}\) and \(b=nk_{2}\) for some \(k_{1},k_{2} \in \mathbb{Z}\). We now see that
\begin{align*}
a-b&=nk_{1}-nk_{2} \\
&=n(k_{1}-k_{2}) \in n \mathbb{Z},
\end{align*}
since \(k_{1}-k_{2} \in \mathbb{Z}\). Hence, \(a-b \in n\mathbb{Z}\) for all \(a,b \in n\mathbb{Z}\), so \(n \mathbb{Z}\) is a subroup of \(\mathbb{Z}\).

Conclusion

Thanks for reading, and I hope this helps you work out when you have a subgroup of a given group. If this was helpful, make sure to check out our other Algebra posts and share with others that may need the help.

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