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Possible Duplicate:
Basic Subgroup Conditions

could someone please explain how the one step subgroup test works,

I know its important and everything but I do not know how to apply it as well as with the two step subgroup.

If someone could also give some examples with it it would be really helpful.

thank you

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marked as duplicate by Gerry Myerson, Brandon Carter, t.b., user31373, Zhen Lin Aug 16 '12 at 10:20

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Rather than prove that the "one step subgroup test" and the "two step subgroup test" are equivalent (which the links in the comments do very well), I thought I would "show it in action".

Suppose we want to show that $2\Bbb Z = \{k \in \Bbb Z: k = 2m, \text{for some }m \in \Bbb Z\}$ is a subgroup of $\Bbb Z$ under addition.

A) The "two-step method": first, we show closure - given $k,k' \in 2\Bbb Z$, we have that:

$k = 2m,k' = 2m'$ for some integers $m,m'$, so $k+k' = 2m+2m' = 2(m+m')$. Since $\Bbb Z$ is a group, and closed under addition, $m+m'$ is an integer, so $k+k' \in 2\Bbb Z$.

Next, we show that if $k \in 2\Bbb Z$, $-k \in 2\Bbb Z$: since $k = 2m$, for some integer $m$, we have $-k = -(2m) = 2(-m)$, and since $-m$ is also an integer, $-k \in 2\Bbb Z$.

B) The "one step method": here, we combine both steps into one: given $k,k' \in 2\Bbb Z$, we aim to show that $k + (-k') \in 2\Bbb Z$. As before, we write:

$k + (-k') = k - k' = 2m - 2m' = 2(m -m')$, and since $m - m'$ is an integer, $k + (-k') \in 2\Bbb Z$.

A more sophisticated use of this test, is to show that for any subgroup $H$ of a group $G$, and any element $g \in G$, $gHg^{-1} = \{ghg^{-1}: h \in H\}$ is also a subgroup of $G$. So given any pair of elements $x,y \in gHg^{-1}$, we must show $xy^{-1} \in gHg^{-1}$. Note we can write:

$x = ghg^{-1}$, for some $h \in H$, $y = gh'g^{-1}$, for some $h'\in H$.

Then $y^{-1} = (gh'g^{-1})^{-1} = (g^{-1})^{-1}h'^{-1}g^{-1} = gh'^{-1}g^{-1}$, so:

$xy^{-1} = (ghg^{-1})(gh'^{-1}g^{-1}) = gh(g^{-1}g)h'^{-1}g^{-1} = gh(e)h'^{-1}g^{-1} = g(hh'^{-1})g^{-1}$.

Since $H$ is a subgroup, it contains all inverses, so $h'^{-1}$ is certainly in $H$, and $H$ is also closed under multiplication, so $hh'^{-1} \in H$, thus:

$xy^{-1} = g(hh'^{-1})g^{-1} \in gHg^{-1}$, and we are done.

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    $\begingroup$ What bothers me about the "one-step" method is that surely one also has to prove that the purported subgroup is non-empty, so it's really 2 steps (and the "two-step" method is really 3 steps). Of course, it's usually obvious that the set isn't empty, as in your example, $2{\bf Z}$, but it still ought to be noted. $\endgroup$ – Gerry Myerson Aug 15 '12 at 4:38
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    $\begingroup$ you guys are awesome that really helped me out thanks again $\endgroup$ – user37012 Aug 15 '12 at 17:04

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