Show that the set $H= \{[0]_{10}, [5]_{10}\}$ is a subgroup of $\Bbb Z_{10}$.

Question

Show that the set $H= \{[0]_{10}, [5]_{10}\}$ is a subgroup of $\Bbb Z_{10}$.

Here is what I have currently. If $a,b \in H$, then since $H$ only contains two elements WLOG assume $a = [0]_{10}$ and $b=[5]_{10}$. Thus $a+b = [0]_{10} + [5]_{10} = [0+5]_{10} \in H$.

Now it seems that the identity is $[0]_{10}$ since for any $[h]_{10}$ we have that $[0]_{10} + [h]_{10} = [h]_{10}$.

I’m a bit stuck on how to show that for every $h \in H$, there exitst $h^{-1} \in H$. What is the inverse even for example for $[5]_{10}$?

Answer 1

Since $H$ is a nonempty, finite subset of $\Bbb Z_{10}$, it suffices to show that $H$ is closed under $+$, the operation of $\Bbb Z_{10}$.

To this end, simply observe the Cayley table:

$$\begin{array}{c|cc} + & [0]_{10} & [5]_{10} \\ \hline [0]_{10} & [0]_{10} & [5]_{10} \\ [5]_{10} & [5]_{10} & [0]_{10}. \end{array}$$

The inverse of $[5]_{10}$ is itself. This can be verified by direct computation.

Answer 2

Let $G=\mathbb{Z}_{10}$ then for every divisor of $10$, $G$ has a subgroup. That is, Subgroups are $\mathbb{Z}_{10}$, $2\mathbb{Z}_{10}$,$5\mathbb{Z}_{10}$ and $10\mathbb{Z}_{10}$

Let $H=5\mathbb{Z}_{10}=\{0,5\}$ under addition modulo $10$, then by 2 step subgroup test;

for all $a,b \in \mathbb{Z}_{10}$ clearly $a+_{10}b \in \mathbb{Z}_{10}$,

for all $a \in \mathbb{Z}_{10}$ , $a^{-1}\in \mathbb{Z}_{10}$, In $H$, $0^{-1}=0 $ in $ \mathbb{Z}_{10}$ and $1^{-1}=1 $ in $ \mathbb{Z}_{10}$ Hence $H$ is a subgroup of $G=\mathbb{Z}_{10}$, and simliary one can proof for other subgroups also.