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Let the additive group of the integers be $(G = Z, +)$. Then $H = nℤ = ${$na;$ $a ∈ ℤ$ } is a subgroup of $G$, with $n ∈ N$

my attempt

To show that H is a subgroup of G, H must have the identity of G, an inverse for each element in the group, and closed under addition.

Identity

Let $ g,h ∈ H$ and $h = g^{-1}$

$g + h = 0$

Inverse

Let $g ∈ H$

Given

$g = na$

Therefore

$g^{-1}$=$na^{-1}$ for all $g ∈ H$

Closed under addition

Let $ g,h ∈ H$

Given

$g =$ $na_1$ $h=na_2$

$g+h =$ $na_1 + na_2$ = $n(a_1+a_2)$

Therefore $g+h ∈ H$ for all $g,h$

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    $\begingroup$ Identity: $0 = 0a \in H$. $\endgroup$ – GAVD Sep 22 '17 at 1:54
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I don't see any mistakes in what you've done so far. Even though it is trivial, be sure to mention a word or two about how associativity of $H$ follows from the original associativity of $G$. Also, be mindful of your notation. $G$ is clearly defined to be a group under addition. Hence, you should use $-h$ to denote the inverse of $h \in H$, not $h^{-1}$. Finally, as @GAVD commented, be sure to check that your identity satisfies $0 + h = h + 0 = h$.

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